RD Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 11 Linear Programming are provided here with simple stepbystep explanations. These solutions for Linear Programming are extremely popular among class 12 Science students for Math Linear Programming Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2018 Book of class 12 Science Math Chapter 11 are provided here for you for free. You will also love the adfree experience on Meritnation’s RD Sharma XII Vol 2 2018 Solutions. All RD Sharma XII Vol 2 2018 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.
Page No 30.14:
Question 1:
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of manhours of labour required in each shop for the production of each unit of A and B, and the number of manhours the firm has available per week are as follows:
Gadget  Foundry  Machineshop 
A  10  5 
B  6  4 
Firm's capacity per week  1000  600 
The profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
Answer:
Let x and y number of gadgets A and B respectively being produced in order to maximize the profit.
Since, each unit of gadget A takes 10 hours to be produced by machine A and 6 hours to be produced by machine B and each unit of gadget B takes 5 hours to be produced by machine A and 4 hours to be produced by machine B.
Therefore, the total time taken by the Foundry to produce x units of gadget A and y units of gadget B is $10x+6y$. This must be less than or equal to the total hours available.
Hence, 10x + 6y ≤ 1000.
This is our first constraint.
The total time taken by the machineshop to produce x units of gadget A and y units of gadget B is 5x + 4y. This must be less than or equal to the total hours available.
Hence, 5x + 4y ≤ 600
This is our second constraint.
Since x and y are non negative integers, therefore $x,y\ge $$0$
It is given that the profit on the sale of A is Rs 30 per unit as compared with Rs 20 per unit of B. Therefore, profit gained on x and y number of gadgets A and B is Rs 30x and Rs 20y respectively.
Let Z denotes the total cost
Therefore, Z= Rs (30x + 20y)
Hence, the above LPP can be stated mathematically as follows:
Maximize Z = 30x + 20y
subject to
10x + 6y ≤ 1000,
5x + 4y ≤ 600
x, y ≥ 0
Page No 30.14:
Question 2:
A company is making two products A and B. The cost of producing one unit of products A and B are Rs 60 and Rs 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate the problem as a LPP.
Answer:
Let the company produces x units of product A and y units of product B.
Since, each unit of product A costs Rs 60 and each unit of product B costs Rs 80.Therefore, x units of product A and y units of product B will cost Rs 60x and Rs 80y respectively.
Let Z denotes the total cost.
$\therefore $ Z = Rs (60x + 80y)
Also, one unit of product A requires one machine hour.
The total machine hours available with the company for product A are 400 hours.
Therefore, $x\le 400$
This is our first constraint
Also,one unit of product A and B require 1 labour hour each and there are a total of 500 labours hours.
Thus, $x+y\le 500$
This is our second constraint.
Since, x and y are non negative integers, therefore $x,y\ge $$0$
Also, as per agreement, the company has to supply atleast 200 units of product B to its regular customers.
$\therefore y\ge 200$
Hence, the required LPP is as follows:
Minimize Z = 60x + 80y
subject to
x $\le $$400$,
x + y $\le $$500$
$y\ge 200$
$x,y\ge $$0$
Page No 30.14:
Question 3:
A firm manufactures 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and below is the required processing time in minutes for each machine on each product:
Machine  Products  
A  B  C  
M_{1} M_{2} 
4  3  5 
2  2  4 
Machines M_{1} and M_{2}_{ }have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's. Set up a LPP to maximize the profit.
Answer:
Let the number of units of product A, B and C manufactured be x, y and z respectively.
Given, machine ${M}_{1}$ takes 4 minutes to manufacture 1 unit of product A, 3 minutes to manufacture one unit of product B and 5 minute to manufacture one unit of product C.
Machine ${M}_{2}$ takes 2 minutes to manufacture 1 unit of product A, 2 minutes to manufacture one unit of product B and 4 minute to manufacture one unit of product C.
The availability is 2000 minutes for ${M}_{1}$ and 2500 minutes for ${M}_{2}$
Thus,
$4x+3y+5z\le 2000\phantom{\rule{0ex}{0ex}}2x+2y+4z\le 2500$
Number of units of products cannot be negative.
So, $x,y,z\ge 0$
Further, it is given that the firm should manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's.
Then,
$100\le x\le 150\phantom{\rule{0ex}{0ex}}y\ge 200\phantom{\rule{0ex}{0ex}}z\ge 50$
Let Z denotes the profit
$\therefore \mathrm{Z}=$3x + 2y + 4z
Hence, the required LPP is as follows :
Maximize $\mathrm{Z}=$3x + 2y + 4z
subject to
$4x+3y+5z\le 2000\phantom{\rule{0ex}{0ex}}2x+2y+4z\le 2500$
$100\le x\le 150\phantom{\rule{0ex}{0ex}}y\ge 200\phantom{\rule{0ex}{0ex}}z\ge 50\phantom{\rule{0ex}{0ex}}x,y,z\ge 0$
Page No 30.15:
Question 4:
A firm manufactures two types of products A and B and sells them at a profit of Rs 2 on type A and Rs 3 on type B. Each product is processed on two machines M_{1} and M_{2}. Type A requires one minute of processing time on M_{1} and two minutes of M_{2}; type B requires one minute on M_{1} and one minute on M_{2}. The machine M_{1} is available for not more than 6 hours 40 minutes while machine M_{2} is available for 10 hours during any working day. Formulate the problem as a LPP.
Answer:
Let the firm produces x units of product A and y units of product B.
Since, each unit of product A requires one minute on machine ${M}_{\mathit{1}}$ and two minutes on machine ${M}_{2}$.
Therefore, x units of product A will require product $x$ minutes on machine ${M}_{\mathit{1}}$ and $2x$ minutes on machine ${M}_{2}$
Also,
Since each unit of product B requires one minute on machine ${M}_{\mathit{1}}$ and one minute on machine ${M}_{2}$.
Therefore, $y$ units of product A will require product $y$ minutes on machine ${M}_{\mathit{1}}$ and $y$ minutes on machine ${M}_{2}$
It is given that the machine ${M}_{1}$ is available for $6\mathrm{hours}\mathrm{and}40\mathrm{minutes}$ i.e. $400\mathrm{minutes}$ and machine ${M}_{2}$ is available for 10 hours i.e. 600 minutes
Thus,
$x+y\le 400\phantom{\rule{0ex}{0ex}}2x+y\le 600$
Since,units of the products cannot be negative,so $x,y\ge $$0$
Let Z denotes the total profit
$\mathit{\therefore}\mathit{}\mathrm{Z}=2x+3y$ which is to be maximised
Hence, the required LPP is as follows:
Maximize Z = 2x + 3y
subject to
$x+y\le 400\phantom{\rule{0ex}{0ex}}2x+y\le 600$
$x,y\ge $$0$
Page No 30.15:
Question 5:
A rubber company is engaged in producing three types of tyres A, B and C. Each type requires processing in two plants, Plant I and Plant II. The capacities of the two plants, in number of tyres per day, are as follows:
Plant  A  B  C 
I  50  100  100 
II  60  60  200 
The monthly demand for tyre A, B and C is 2500, 3000 and 7000 respectively. If plant I costs Rs 2500 per day, and plant II costs Rs 3500 per day to operate, how many days should each be run per month to minimize cost while meeting the demand? Formulate the problem as LPP.
Answer:
Let plant I be run for x days and plant II be run for y days
Then,
Tyres  Plant I (x)  Plant II (y)  Demand 
A  50  60  2500 
B  100  60  3000 
C  100  200  7000 
Minimum demand for Tyres A,B and C is 2500, 3000 and 7000 respectively.The demand can be more than the minimum demand.
Therefore,the inequations will be
$50x+60y\ge 2500\phantom{\rule{0ex}{0ex}}100x+60y\ge 3000$
$100x+200y\ge 7000$
Also, the objective function is Z = 2500x + 3500y
Hence, the required LPP is as follows:
Minimise Z = 2500x + 3500y
subject to
$50x+60y\ge 2500\phantom{\rule{0ex}{0ex}}100x+60y\ge 3000$
$100x+200y\ge 7000$
Page No 30.15:
Question 6:
A company sells two different products A and B. The two products are produced in a common production process and are sold in two different markets. The production process has a total capacity of 45000 manhours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of units of A that can be sold is 7000 and that of B is 10,000. If the profit is Rs 60 per unit for the product A and Rs 40 per unit for the product B, how many units of each product should be sold to maximize profit? Formulate the problem as LPP.
Answer:
Let x units of product A and y units of product B be produced.
Then,
Since, it takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.
Therefore, it will take 5x hours to produce x units of A and 3y hours to produce y units of B.
As, the total capacity is of 45000 man hours.
$\Rightarrow 5x+3y\le 45000$
Also,
The maximum number of units of A that can be sold is 7000 and that of B is 10,000 and number of units cannot be negative.
Thus, $0\le x\le 7000,0\le y\le 10000$
Now,
Total profit = $60x+40y$
Here, we need to maximize profit
Thus, the objective function will be maximize $\mathrm{Z}=60x+40y$
Hence, the required LPP is as follows:
Maximize Z = 60x + 40y
subject to
$5x+3y\le 45000\phantom{\rule{0ex}{0ex}}x\le 7000\phantom{\rule{0ex}{0ex}}y\le 10000\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Page No 30.15:
Question 7:
To maintain his health a person must fulfil certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrientscalcium, protein and calories and the person's diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:
Food I (per lb) 
Food II (per lb) 
Minimum daily requirement for the nutrient 

Calcium  10  5  20  
Protein  5  4  20  
Calories  2  6  13  
Price (Rs)  60  100 
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.
Answer:
Let the person takes x lbs and y lbs of food I and II respectively that were taken in the diet.
Since, per lb of food I costs Rs 60 and that of food II costs Rs 100.
Therefore, x lbs of food I costs Rs 60x and y lbs of food II costs Rs 100y.
Total cost per day = Rs (60x + 100y)
Let Z denote the total cost per day
Then, Z = 60x + 100y
Total amount of calcium in the diet is $10x+5y$
Since, each lb of food I contains 10 units of calcium.Therefore, x lbs of food I contains 10x units of calcium.
Each lb of food II contains 5 units of calciu.So,y lbs of food II contains 5y units of calcium.
Thus, x lbs of food I and y lbs of food II contains 10x + 5y units of calcium.
But, the minimum requirement is 20 lbs of calcium.
$\therefore $ $10x+5y\ge 20$
Since, each lb of food I contains 5 units of protein.Therefore, x lbs of food I contains 5x units of protein.
Each lb of food II contains 4 units of protein.So,y lbs of food II contains 4y units of protein.
Thus, x lbs of food I and y lbs of food II contains 5x + 4y units of protein.
But, the minimum requirement is 20 lbs of protein.
$\therefore $ $5x+4y\ge 20$
Since, each lb of food I contains 2 units of calories.Therefore, x lbs of food I contains 2x units of calories.
Each lb of food II contains units of calories.So,y lbs of food II contains 6y units of calories.
Thus, x lbs of food I and y lbs of food II contains 2x + 6y units of calories.
But, the minimum requirement is 13 lbs of calories.
$\therefore 2x+6y\ge 13$
Finally, the quantities of food I and food II are non negative values.
So, $x,y\ge 0$
Hence, the required LPP is as follows:
Min Z = 60x + 100y
subject to
$10x+5y\ge 20\phantom{\rule{0ex}{0ex}}5x+4y\ge 20\phantom{\rule{0ex}{0ex}}2x+6y\ge 13\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Page No 30.16:
Question 8:
A manufacturer can produce two products, A and B, during a given time period. Each of these products requires four different manufacturing operations: grinding, turning, assembling and testing. The manufacturing requirements in hours per unit of products A and B are given below.
A  B  
Grinding  1  2 
Turning  3  1 
Assembling  6  3 
Testing  5  4 
The available capacities of these operations in hours for the given time period are: grinding 30; turning 60, assembling 200; testing 200. The contribution to profit is Rs 20 for each unit of A and Rs 30 for each unit of B. The firm can sell all that it produces at the prevailing market price. Determine the optimum amount of A and B to produce during the given time period. Formulate this as a LPP.
Answer:
Let x and y units of products A and B were manufactured respectively.
The contribution to profit is Rs 2 for each unit of A and Rs 3 for each unit of B.
Therefore for x units of A and y units of B,the contribution to profit would be Rs 2x and Rs 3y respectively.
Let Z denote the total profit
Then, Z = Rs (2x + 3y)
Total hours required for grinding, turning, assembling and testing are $x+2y,3x+y,6x+3y,5x+4y$ respectively.
The available capacities of these operations in hours for the given period are grinding 30, turning 60, assembling 200 and testing 200.
$\therefore $ $x+2y\le 30,3x+y\le 60,6x+3y\le 200,5x+4y\le 200$
Units of products cannot be negative.Therefore,
$x,y\ge 0$
Hence, the required LPP is as follows:
Maximize Z = 2x + 3y
subject to
$x+2y\le 30,\phantom{\rule{0ex}{0ex}}3x+y\le 60,\phantom{\rule{0ex}{0ex}}6x+3y\le 200,\phantom{\rule{0ex}{0ex}}5x+4y\le 200$
Page No 30.16:
Question 9:
Vitamins A and B are found in two different foods F_{1} and F_{2}. One unit of food F_{1} contains 2 units of vitamin A and 3 units of vitamin B. One unit of food F_{2} contains 4 units of vitamin A and 2 units of vitamin B. One unit of food F_{1} and F_{2} cost Rs 50 and 25 respectively. The minimum daily requirements for a person of vitamin A and B is 40 and 50 units respectively. Assuming that any thing in excess of daily minimum requirement of vitamin A and B is not harmful, find out the optimum mixture of food F_{1} and F_{2} at the minimum cost which meets the daily minimum requirement of vitamin A and B. Formulate this as a LPP.
Answer:
Let x and y units of food F_{1} and food F_{2} were mixed.
Clearly, x ≥ 0 and y ≥ 0
One unit of food F_{1} contains 2 units of vitamin A and one unit of of food F_{2} contains 4 units of vitamin A. Therefore, x and y units of food F_{1} and food F_{2} respectively contains 2x and 4y units of vitamin A.
It is given that the minimum daily requirements for a person of vitamin A is 40 units.
Hence, 2x+ 4y ≥ 40
One unit of food F_{1} contains 3 units of vitamin B and one unit of food F_{2} contains 2 units of of vitamin B. Therefore, x and y units of F_{1} and F_{2} respectively contains 3x and 2y units of vitamin B.
It is given that the minimum daily requirements for a person of vitamin B is 50 units.
Hence, 3x+ 2y ≥ 50
One unit of food F_{1} and food F_{2} cost Rs 50 and 25 respectively. Therefore, x and y units of food F_{1} and food F_{2} costs Rs 50x and Rs 25y respectively.
Let Z denote the total cost
Then, Z = Rs (50x + 25y)
Hence, the required LPP is
Minimize Z = 50x + 25y
subject to
2x+ 4y ≥ 40
3x+ 2y ≥ 50
x ≥ 0,y ≥ 0
Page No 30.16:
Question 10:
An automobile manufacturer makes automobiles and trucks in a factory that is divided into two shops. Shop A, which performs the basic assembly operation, must work 5 mandays on each truck but only 2 mandays on each automobile. Shop B, which performs finishing operations, must work 3 mandays for each automobile or truck that it produces. Because of men and machine limitations, shop A has 180 mandays per week available while shop B has 135 mandays per week. If the manufacturer makes a profit of Rs 30000 on each truck and Rs 2000 on each automobile, how many of each should he produce to maximize his profit? Formulate this as a LPP.
Answer:
Let x number of trucks and y number of automobiles were produced to maximize the profit.
Since, the manufacturer makes profit of Rs 30000 on each truck and Rs 2000 on each automobile.
Therefore, on x number of trucks and y number of automobiles profit would be Rs 30000x and Rs 2000y respectively.
Total profit = Rs (30000x + 2000y)
Let Z denote the total profit
Then, Z = 30000x + 2000y
Since, 5 mandays and 2 mandays were required to produce each truck and automobile at shop A.
Therefore, 5x mandays and 2y mandays are required to produce x trucks and y automobiles at shop A.
Also,
Since 3 mandays were required to produce each truck and automobile at shop B.
Therefore, 3x mandays and 3y mandays are required to produce x trucks and y automobiles.
As, shop A has 180 mandays per week available while shop B has 135 mandays per week.
$\therefore $ $5x+2y\le 180,3x+3y\le 135$
Number of trucks and automobiles cannot be negative.
$\therefore $ $x,y\ge 0$
Hence, the required LPP is as follows :
Maximize Z = 30000x + 2000y
subject to
$5x+2y\le 180,\phantom{\rule{0ex}{0ex}}3x+3y\le 135,\phantom{\rule{0ex}{0ex}}x\ge 0,y\ge 0$
Page No 30.16:
Question 11:
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
Product A  Product B  Weekly capacity  
Department 1  3  2  130 
Department 2  4  6  260 
Selling price per unit  Rs 25  Rs 30  
Labour cost per unit  Rs 16  Rs 20  
Raw material cost per unit  Rs 4  Rs 4 
The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.
Answer:
Let x and y units of product A and B were manufactured respectively.
Labour cost per unit to manufacture product A and product B is Rs 16 and Rs 20 respectively.Therefore, labour cost for x and y units of product A and product B is Rs 16x and Rs 20y respectively.
Total labour cost to manufacture product A and product B is Rs (16x+20y)
Raw material cost per unit to manufacture product A and product B is Rs 4 and Rs 4 respectively.Therefore,raw material cost for x and y units of product A and product B is Rs 4x and Rs 4y respectively.
Total raw material cost to manufacture product A and product B is Rs (4x + 4y)
Hence, total cost price to manufacture product A and product B = Total labour cost + Total raw material cost
= 16x + 4x + 20y + 4y
= 20x + 24y
Selling price per unit for product A and product B is Rs 25 and Rs 30 respectively. Therefore, total selling price for product A and product B is Rs 25x and Rs 30y respectively.
Total selling price = 25x + 30y
∴ Total profit = Total selling price − Total cost price = 25x + 30y $$(20x + 24y)
= 5x + 6y
Let Z denote the total profit
Then, Z = $5x+6y$
One unit of product A and product B requires 3 hours and 2 hours respectively at department 1.Therefore, x units and y units of product A and product B
require 3x hours and 2y hours respectively.
The weekly capacity of department 1 is 130.
$\therefore 3x+2y\le 130$
One unit of product A and B requires 4 hours and 6 hours respectively at department 2.Therefore, x units and y units of product A and product B require 4x hours and 6y hours respectively.
The weekly capacity of department 2 is 260.
$\therefore 4x+6y\le 260$
Units of products cannot be negative.Therefore,
$x,y\ge 0$
Hence, the required LPP is as follows:
Maximize Z = 5x + 6y
subject to
$3x+2y\le 130,\phantom{\rule{0ex}{0ex}}4x+6y\le 260,\phantom{\rule{0ex}{0ex}}x\ge 0,y\ge 0$
Page No 30.16:
Question 12:
An airline agrees to charter planes for a group. The group needs at least 160 first class seats and at least 300 tourist class seats. The airline must use at least two of its model 314 planes which have 20 first class and 30 tourist class seats. The airline will also use some of its model 535 planes which have 20 first class seats and 60 tourist class seats. Each flight of a model 314 plane costs the company Rs 100,000 and each flight of a model 535 plane costs Rs 150,000. How many of each type of plane should be used to minimize the flight cost? Formulate this as a LPP.
Answer:
Let $x$ number of model 314 planes and y number of model 535 planes were used.
It is given that cost of one model 314 plane is Rs 100000 and cost of one model 535 plane is Rs 150000.
Therefore, cost of x model 314 plane is Rs 100000x and cost of y model 535 plane is Rs 150000y.
Total cost price = 100000x + 150000y
Let Z denote the total cost
Then, Z = $100000x+150000y$
Also,
Each model 314 planes have 20 first class and 30 tourist class seats and each model 535 planes has 20 first class and 60 tourist class seats.
The group needs 160 first class seats and 300 tourist class seats.
$\therefore 20x+20y\ge 160\phantom{\rule{0ex}{0ex}}30x+60y\ge 300$
Number of planes cannot be negative.
Therefore, $x,y\ge 0$
Min Z = $100000x+150000y$
subject to
$20x+20y\ge 160\phantom{\rule{0ex}{0ex}}30x+60y\ge 300\phantom{\rule{0ex}{0ex}}x\ge 0,y\ge 0$
Page No 30.16:
Question 13:
Amit's mathematics teacher has given him three very long lists of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he can not afford to devote more than $3\frac{1}{2}$ hours altogether to his mathematics assignment. Moreover, the first two sets of problems involve numerical calculations and he knows that he cannot stand more than $2\frac{1}{2}$ hours work on this type of problem. Under these circumstances, how many problems in each of these categories shall he do in order to get maximum possible credit for his efforts? Formulate this as a LPP.
Answer:
Let Amit correctly solves x problems from the first set, y problems from the second set and z problems from the third set.
Given,
Amit cannot submit more than 100 correctly solved problems.
$\therefore x+y+z\le 100$
The problem in the first set are worth 5 points each,those in the second set worth 4 points each and those in the third set worth 6 points each.
Therefore, x problems from the first set worth 5x points, y problems from the second set worth 4y points and z problems from the third set worth 6z points.
Thus, total credit points will be $5x+4y+6z$.
Let Z denotes the total credit of Amit
$\therefore \mathrm{Z}=5x+4y+6z$
It requires 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem and 4 minutes to solve a 6 point problem.Therefore,x problems from the first set require 3x minutes, y problems from the second set require 2y minutes and z problems from the third set require 4z minutes.
Thus, the total time require by Amit will be (3x + 2y + 4z) minutes.
It is given that the total time that Amit can devote on his mathematics assignment is $3\frac{1}{2}\mathrm{hours}\mathrm{i}.\mathrm{e}.210\mathrm{minutes}.$
$\therefore 3x+2y+4z\le 210$
Further, it is given that the total time that Amit can devote in solving first two types of problems cannot be more than $2\frac{1}{2}$ hours i.e. $150\mathrm{minutes}$.
$\therefore 3x+2y\le 150$
Number of problems cannot be negative.Therefore,
$x,y\ge 0$
Maximize $\mathrm{Z}=5x+4y+6z$
subject to
$x+y+z\le 100\phantom{\rule{0ex}{0ex}}3x+2y+4z\le 210\phantom{\rule{0ex}{0ex}}3x+2y\le 150\phantom{\rule{0ex}{0ex}}x\ge 0,y\ge 0$
Page No 30.17:
Question 14:
A farmer has a 100 acre farm. He can sell the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes, 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is available at Rs 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 mandays for tomatoes and radishes and 6 mandays for lettuce. A total of 400 mandays of labour are available at Rs 20 per manday. Formulate this problem as a LPP to maximize the farmer's total profit.
Answer:
Let the farmer sow tomatoes in x acres, lettuce in y acres & radishes in z acres of the farm.
Average yield per acre is 2000 kgs for tomatoes, 3000 kgs of lettuce and 1000 kg of radishes.
Thus, the farmer raised 2000x kg of tomatoes, 3000y kg of lettuce and 1000z kg of radishes.
Given, price he can obtain is Re 1 per kilogram for tomatoes, Re 0.75 a head for lettuce and Rs 2 per kilogram for radishes.
∴ Selling price = Rs $\left[2000x\left(1\right)+3000y\left(0.75\right)+1000z\left(2\right)\right]$ = Rs (2000x + 2250y + 2000z)
Labour required for sowing, cultvating and harvesting per acre is 5 mandays for tomatoes and radishes and 6 mandays for lettuce.Therefore, labour required for sowing, cultivating and harvesting per acre is 5x for tomatoes, 6y for lettuce and 5z for radishes.
Number of mandays required in sowing, cultivating and harvesting= $5x+6y+5z$
Price of one manday = Rs 20
$\therefore $ Labour cost = $20\left(5x+6y+5z\right)=100x+120y+100z$
Also, fertilizer is available at Re 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kgs for radishes.
Therefore, fertilizer required is 100x kgs for the tomatoes sown in x acres, 100y kgs for the lettuce sown in y acres and 50z kgs for radishes sown in z acres of land.
Hence, total fertilizer used= (100x + 100y +50z) kgs
Thus, fertilizer's cost = $\mathrm{Rs}0.5\times \left(100x+100y+50z\right)=\mathrm{Rs}\left(50x+50y+25z\right)$
So, the total price that has been cost to farmer = Labour cost + Fertilizer cost
= Rs $\left(150x+170y+125z\right)$
Profit made by farmer = Selling price $$ Cost price
= Rs (2000x + 2250y + 2000z)− Rs (150x + 170y + 125z)
= Rs $\left(1850x+2080y+1875z\right)$
Let Z denotes the total profit
$\therefore \mathrm{Z}=1850x+2080y+1875z$
Now,
Total area of the farm = 100 acres
$x+y+z\le 100$
Also, it is given that the total mandays available are 400.
Thus, $5x+6y+5z\le 400$
Area of the land cannot be negative.
Therefore, $x,y\ge 0$
Maximize $Z=1850x+2080y+1875z$
subject to
$x+y+z\le 100\phantom{\rule{0ex}{0ex}}5x+6y+5z\le 400\phantom{\rule{0ex}{0ex}}x,y,z\ge 0$
Page No 30.17:
Question 15:
A firm has to transport at least 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost of engaging each large van is ₹400 and each small van is ₹200. Not more than ₹3000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost.
Answer:
Let the number of large vans and small vans used for transporting the packages be x and y, respectively.
It is given that the cost of engaging each large van is ₹400 and each small van is ₹200.
Cost of engaging x large vans = ₹400x
Cost of engaging y small vans = ₹200y
Let Z be the total cost of engaging x large vans and y small vans.
∴ Z = ₹(400x + 200y)
The firm has to transport at least 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each.
∴ Number of packages transported by x large vehicles + Number of packages transported by y small vehicles ≥ 1200
⇒ 200x + 80y ≥ 1200
Not more than ₹3000 is to be spent daily on the transportation.
∴ 400x + 200y ≤ 3000
Also, the number of large vans cannot exceed the number of small vans.
∴ x ≤ y
Thus, the linear programming problem of the given problem is
Minimise Z = ₹(400x + 200y)
Subject to constraints
200x + 80y ≥ 1200
400x + 200y ≤ 3000
x ≤ y
x ≥ 0, y ≥ 0
Page No 30.32:
Question 1:
Maximize Z = 5x + 3y
Subject to
$3x+5y\le 15\phantom{\rule{0ex}{0ex}}5x+2y\le 10\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
Region represented by 3x + 5y ≤ 15 :
The line 3x + 5y = 15 meets the coordinate axes at A(5,0) and B(0,3) respectively. By joining these points we obtain the line 3x + 5y = 15.
Clearly (0,0) satisfies the inequation 3x + 5y ≤ 15. So,the region containing the origin represents the solution set of the inequation 3x + 5y ≤ 15.
Region represented by 5x + 2y ≤ 10 :
The line 5x + 2y = 10 meets the coordinate axes at C(2,0) and D(0,5) respectively. By joining these points we obtain the line 5x + 2y = 10.
Clearly (0,0) satisfies the inequation 5x + 2y ≤ 10. So,the region containing the origin represents the solution set of the inequation 5x + 2y ≤ 10.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), C(2, 0),$E\left(\frac{20}{19},\frac{45}{19}\right)$ and B(0, 3).
The values of Z at these corner points are as follows.
Corner point  Z = 5x + 3y 
O(0, 0)  5 × 0 + 3 × 0 = 0 
C(2, 0)  5 × 2 + 3 × 0 = 10 
$E\left(\frac{20}{19},\frac{45}{19}\right)$  5 × $\frac{20}{19}$ + 3 × $\frac{45}{19}$ = $\frac{235}{19}$ 
B(0, 3)  5 × 0 + 3 × 3 = 9 
Therefore, the maximum value of Z is $\frac{235}{19}\mathrm{at}\mathrm{the}\mathrm{point}\left(\frac{20}{19},\frac{45}{19}\right)$.Hence, x= $\frac{20}{19}$ and y =$\frac{45}{19}$ is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{235}{19}$.
Page No 30.32:
Question 2:
Maximize Z = 9x + 3y
Subject to
$2x+3y\le 13\phantom{\rule{0ex}{0ex}}3x+y\le 5\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
2x + 3y = 13, 3x +y = 5, x = 0 and y = 0
Region represented by 2x + 3y ≤ 13 :
The line 2x + 3y = 13 meets the coordinate axes at $A\left(\frac{13}{2},0\right)$ and $B\left(0,\frac{13}{3}\right)$ respectively. By joining these points we obtain the line 2x + 3y = 13.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 13. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 13.
Region represented by 3x + y ≤ 5:
The line 5x + 2y = 10 meets the coordinate axes at $C\left(\frac{5}{3},0\right)$ and D(0, 5) respectively. By joining these points we obtain the line 3x + y = 5.
Clearly (0,0) satisfies the inequation 3x + y ≤ 5. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 5.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints, 2x + 3y ≤ 13, 3x + y ≤ 5, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), $C\left(\frac{5}{3},0\right)$ ,$E\left(\frac{2}{7},\frac{29}{7}\right)$ and $B\left(0,\frac{13}{3}\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = 9x + 3y 
O(0, 0)  9 × 0 + 3 × 0 = 0 
$C\left(\frac{5}{3},0\right)$  9 × $\frac{5}{3}$ + 3 × 0 = 15 
$E\left(\frac{2}{7},\frac{29}{7}\right)$  9 × $\frac{2}{7}$ + 3 × $\frac{29}{7}$ = 15 
$B\left(0,\frac{13}{3}\right)$  9 × 0 + 3 ×$\frac{13}{3}$ = 13 
We see that the maximum value of the objective function Z is 15 which is at C $\left(\frac{5}{3},0\right)$ and $E\left(\frac{2}{7},\frac{29}{7}\right)$.
Thus, the optimal value of Z is 15.
Page No 30.32:
Question 3:
Minimize Z = 18x + 10y
Subject to
$4x+y\ge 20\phantom{\rule{0ex}{0ex}}2x+3y\ge 30\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
4x + y = 20, 2x +3y = 30, x = 0 and y = 0
Region represented by 4x + y ≥ 20 :
The line 4x + y = 20 meets the coordinate axes at A(5, 0) and B(0, 20) respectively. By joining these points we obtain the line 4x + y = 20.
Clearly (0,0) does not satisfies the inequation 4x + y ≥ 20. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 4x + y ≥ 20.
Region represented by 2x +3y ≥ 30:
The line 2x +3y = 30 meets the coordinate axes at C(15,0) and D(0, 10) respectively. By joining these points we obtain the line
2x +3y = 30.Clearly (0,0) does not satisfies the inequation 2x +3y ≥ 30. So,the region which does not contain the origin represents the solution set of the inequation 2x +3y ≥ 30.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints, 4x + y ≥ 20, 2x +3y ≥ 30, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are B(0, 20), C(15,0), E(3,8) and C(15,0).
The values of Z at these corner points are as follows.
Corner point  Z = 18x + 10y 
B(0, 20)  18 × 0 + 10 × 20 = 200 
E(3,8)  18 × 3 + 10 × 8 = 134 
C(15,0)  18 × 15 + 10 ×0 = 270 
Therefore, the minimum value of Z is 134 at the point E(3,8). Hence, x = 3 and y =8 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 134.
Page No 30.32:
Question 4:
Maximize Z = 50x + 30y
Subject to
$2x+y\le 18\phantom{\rule{0ex}{0ex}}3x+2y\le 34\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 18, 3x + 2y = 34
Region represented by 2x + y ≥ 18:
The line 2x + y = 18 meets the coordinate axes at A(9, 0) and B(0, 18) respectively. By joining these points we obtain the line 2x + y = 18.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 18. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 18.
Region represented by 3x + 2y ≤ 34:
The line 3x + 2y = 34 meets the coordinate axes at $C\left(\frac{34}{3},0\right)$ and D(0, 17) respectively. By joining these points we obtain the line 3x + 2y = 34.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 34. So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 34.
The corner points of the feasible region are A(9, 0), $C\left(\frac{34}{3},0\right)$ and E(2, 14).
The values of Z at these corner points are as follows.
Corner point  Z = 50x + 30y 
A(9, 0)  50 × 9 + 3 × 0 = 450 
$C\left(\frac{34}{3},0\right)$  50 ×$\frac{34}{3}$ + 30 × 0 = $\frac{1700}{3}$ 
E(2, 14)  50 × 2 + 30 × 14 = 520 
Therefore, the maximum value of Z is $\frac{1700}{3}\mathrm{at}\mathrm{the}\mathrm{point}\left(\frac{34}{3},0\right)$.Hence, x = $\frac{34}{3}$ and y =0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{1700}{3}$.
Page No 30.32:
Question 5:
Maximize Z = 4x + 3y
subject to
$3x+4y\le 24\phantom{\rule{0ex}{0ex}}8x+6y\le 48\phantom{\rule{0ex}{0ex}}x\le 5\phantom{\rule{0ex}{0ex}}y\le 6\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
We need to maximize Z = 4x + 3y
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6, x = 0 and y = 0.
The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0,6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24.So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0,8). Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation 8x + 6y ≤ 48. So, the region in xyplane that contains the origin represents the solution set of the given equation.
x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are O(0, 0), $G\left(5,0\right)$, $F\left(5,\frac{4}{3}\right)$, $E\left(\frac{24}{7},\frac{24}{7}\right)$ and $B\left(0,6\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = 4x + 3y 
O(0, 0)  4× 0 + 3 × 0 = 0 
$G\left(5,0\right)$  4 × 5 + 3 × 0 = 20 
$F\left(5,\frac{4}{3}\right)$  4 × 5 + 3 ×$\frac{4}{3}$ = 24 
$E\left(\frac{24}{7},\frac{24}{7}\right)$  4 × $\frac{24}{7}$ + 3 × $\frac{24}{7}$ = $\frac{196}{7}$= 24 
$B\left(0,6\right)$  4 × 0 + 3 × 6 = 18 
We see that the maximum value of the objective function Z is 24 which is at $F\left(5,\frac{4}{3}\right)$ and $E\left(\frac{24}{7},\frac{24}{7}\right)$.
Thus, the optimal value of Z is 24.
Page No 30.32:
Question 6:
Maximize Z = 15x + 10y
Subject to
$3x+2y\le 80\phantom{\rule{0ex}{0ex}}2x+3y\le 70\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0
Region represented by 3x + 2y ≤ 80 :
The line 3x + 2y = 80 meets the coordinate axes at $A\left(\frac{80}{3},0\right)$ and $B\left(0,40\right)$ respectively. By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 80 . So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 80 .
Region represented by 2x + 3y ≤ 70:
The line 2x + 3y = 70 meets the coordinate axes at $C\left(35,0\right)$ and $D\left(0,\frac{70}{3}\right)$ respectively. By joining these points we obtain the line 2x + 3y ≤ 70.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 70. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 70.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.
The corner points of the feasible region are O(0, 0), $A\left(\frac{80}{3},0\right)$ ,$E\left(20,10\right)$ and $D\left(0,\frac{70}{3}\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = 15x + 10y 
O(0, 0)  15 × 0 + 10 × 0 = 0 
$A\left(\frac{80}{3},0\right)$  15 × $\frac{80}{3}$ + 10 × 0 = 400 
$E\left(20,10\right)$  15 × 20 + 10 × 10 = 400 
$D\left(0,\frac{70}{3}\right)$  15 × 0 + 10 ×$\frac{70}{3}$ = $\frac{700}{3}$ 
We see that the maximum value of the objective function Z is 400 which is at $A\left(\frac{80}{3},0\right)$ and $E\left(20,10\right)$.
Thus, the optimal value of Z is 400.
Page No 30.32:
Question 7:
Maximize Z = 10x + 6y
Subject to
$3x+y\le 12\phantom{\rule{0ex}{0ex}}2x+5y\le 34\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
3x + y = 12, 2x + 5y = 34, x = 0 and y = 0
Region represented by 3x + y ≤ 12:
The line 3x + y = 12 meets the coordinate axes at $A\left(4,0\right)$ and $B\left(0,12\right)$ respectively. By joining these points we obtain the line 3x + y = 12.
Clearly (0,0) satisfies the inequation 3x + y ≤ 12. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 12 .
Region represented by 2x + 5y ≤ 34:
The line 2x + 5y = 34 meets the coordinate axes at $C\left(17,0\right)$ and $D\left(0,\frac{34}{5}\right)$ respectively. By joining these points we obtain the line 2x + 5y ≤ 34.
Clearly (0,0) satisfies the inequation 2x + 5y ≤ 34. So,the region containing the origin represents the solution set of the inequation 2x + 5y ≤ 34.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, and y ≥ 0 are as follows.
The corner points of the feasible region are O(0, 0), $A\left(4,0\right)$ ,$E\left(2,6\right)$ and $D\left(0,\frac{34}{5}\right)$.
The values of Z at these corner points are as follows:
Corner point  Z = 10x + 6y 
O(0, 0)  10 × 0 + 6 × 0 = 0 
$A\left(4,0\right)$  10× 4 + 6 × 0 = 40 
$E\left(2,6\right)$  10 × 2 + 6 × 6 = 56 
$D\left(0,\frac{34}{5}\right)$  10 × 0 + 6 ×$\frac{34}{5}$ = $\frac{204}{3}$ 
We see that the maximum value of the objective function Z is 56 which is at $E\left(2,6\right)$ that means at x = 2 and y = 6.
Thus, the optimal value of Z is 56.
Page No 30.32:
Question 8:
Maximize Z = 3x + 4y
Subject to
$2x+2y\le 80\phantom{\rule{0ex}{0ex}}2x+4y\le 120$
Answer:
We have to maximize Z = 3x + 4y
First, we will convert the given inequations into equations, we obtain the following equations:
2x + 2y = 80, 2x + 4y = 120
Region represented by 2x + 2y ≤ 80:
The line 2x + 2y = 80 meets the coordinate axes at $A\left(40,0\right)$ and $B\left(0,40\right)$ respectively. By joining these points we obtain the line 2x + 2y = 80.
Clearly (0,0) satisfies the inequation 2x + 2y ≤ 80. So,the region containing the origin represents the solution set of the inequation 2x + 2y ≤ 80.
Region represented by 2x + 4y ≤ 120:
The line 2x + 4y = 120 meets the coordinate axes at $C\left(60,0\right)$ and $D\left(0,30\right)$ respectively. By joining these points we obtain the line 2x + 4y ≤ 120.
Clearly (0,0) satisfies the inequation 2x + 4y ≤ 120. So,the region containing the origin represents the solution set of the inequation 2x + 4y ≤ 120.
The feasible region determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:
The corner points of the feasible region are O(0, 0), $A\left(40,0\right)$, $E\left(20,20\right)$ and $D\left(0,30\right)$.
The values of Z at these corner points are as follows:
Corner point  Z = 3x + 4y 
O(0, 0)  3 × 0 + 4 × 0 = 0 
$A\left(40,0\right)$  3× 40 + 4 × 0 = 120 
$E\left(20,20\right)$  3 × 20 + 4 × 20 = 140 
$D\left(0,30\right)$  10 × 0 + 4 ×30 = 120 
We see that the maximum value of the objective function Z is 140 which is at $E\left(20,20\right)$ that means at x = 20 and y = 20.
Thus, the optimal value of Z is 140.
Page No 30.32:
Question 9:
Maximize Z = 7x + 10y
Subject to
$x+y\le 30000\phantom{\rule{0ex}{0ex}}y\le 12000\phantom{\rule{0ex}{0ex}}x\ge 6000\phantom{\rule{0ex}{0ex}}x\ge y\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
We have to maximize Z = 7x + 10y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 30000,y = 12000, x = 6000, x = y, x = 0 and y = 0.
Region represented by x + y ≤ 30000:
The line x + y = 30000 meets the coordinate axes at $A\left(30000,0\right)$ and $B\left(0,30000\right)$ respectively. By joining these points we obtain the line x + y = 30000.
Clearly (0,0) satisfies the inequation x + y ≤ 30000. So,the region containing the origin represents the solution set of the inequation x + y ≤ 30000.
The line y = 12000 is the line that passes through C(0,12000) and parallel to x axis.
The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.
Region represented by x ≥ y
The line x = y is the line that passes through origin.The points to the right of the line x = y satisfy the inequation x ≥ y.
Like by taking the point (−12000, 6000).Here, 6000 > −12000 which implies y > x. Hence, the points to the left of the line x = y will not satisfy the given inequation x ≥ y.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, x + y ≤ 30000, y ≤ 12000, x ≥ 6000, x ≥ y , x ≥ 0 and y ≥ 0 are as follows:
The corner points of the feasible region are D(6000, 0), $A\left(3000,0\right)$, $F\left(18000,12000\right)$ and $E\left(12000,12000\right)$.
The values of Z at these corner points are as follows:
Corner point  Z = 7x + 10y 
D(6000, 0)  7 × 6000 + 10 × 0 = 42000 
$A\left(3000,0\right)$  7× 3000 + 10 × 0 = 21000 
$F\left(18000,12000\right)$  7 × 18000 + 10 × 12000 = 246000 
$E\left(12000,12000\right)$  7 × 12000 + 10 ×12000 = 204000 
We see that the maximum value of the objective function Z is 246000 which is at $F\left(18000,12000\right)$ that means at x = 18000 and y = 12000.
Thus, the optimal value of Z is 246000.
Page No 30.32:
Question 10:
Minimize Z = 2x + 4y
Subject to
$x+y\ge 8\phantom{\rule{0ex}{0ex}}x+4y\ge 12\phantom{\rule{0ex}{0ex}}x\ge 3,y\ge 2$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 3, y = 2
Region represented by x + y ≥ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.
Clearly (0,0) does not satisfies the inequation x + y ≥ 8. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 8.
Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.
Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which contain the origin represents the solution set of the inequation x + 4y ≥ 12.
The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis.x ≥ 3 is the region to the right of the line x = 3.
The line y = 2 is the line that passes through the point (0, 12) and is parallel to X axis.y ≥ 2 is the region above the line y = 2.
The corner points of the feasible region are E(3, 5) and F(6, 2).
The values of Z at these corner points are as follows.
Corner point  Z = 2x + 4y 
E(3, 5)  2 × 3 + 4 × 5 = 26 
F(6, 2)  2 × 6 + 4 × 2 = 20 
Therefore, the minimum value of Z is 20 at the point F(6, 2). Hence, x = 6 and y =2 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 20.
Page No 30.32:
Question 11:
Minimize Z = 5x + 3y
Subject to
$2x+y\ge 10\phantom{\rule{0ex}{0ex}}x+3y\ge 15\phantom{\rule{0ex}{0ex}}x\le 10\phantom{\rule{0ex}{0ex}}y\le 8\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 10, x + 3y = 15, x = 10, y = 8
Region represented by 2x + y ≥ 10:
The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively. By joining these points we obtain the line 2x + y = 10.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 10.
Region represented by x + 3y ≥ 15:
The line x + 3y = 15 meets the coordinate axes at C(15, 0) and D(0, 5) respectively. By joining these points we obtain the line x + 3y = 15.
Clearly (0,0) satisfies the inequation x + 3y ≥ 15. o,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 3y ≥ 15.
The line x = 10 is the line that passes through the point (10, 0) and is parallel to Y axis.x ≤ 10 is the region to the left of the line x = 10.
The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis.y ≤ 8 is the region below the line y = 8.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, 2x + y ≥ 10, x + 3y ≥ 15, x ≤ 10, y ≤ 8, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E(3, 4),$H\left(10,\frac{5}{3}\right)$, F(10, 8) and G(1, 8).
The values of Z at these corner points are as follows.
Corner point  Z = 5x + 3y 
E(3, 4)  5 × 3 + 3 × 4 = 27 
$H\left(10,\frac{5}{3}\right)$  5 × 10 + 3× $\frac{5}{3}$ = 55 
F(10, 8)  5 × 10 + 3 × 8 = 74 
G(1, 8)  5 × 1 + 3 × 8 = 29 
Therefore, the minimum value of Z is 27 at the point F(3, 4). Hence, x = 3 and y =4 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 27.
Page No 30.32:
Question 12:
Minimize Z = 30x + 20y
Subject to
$x+y\le 8\phantom{\rule{0ex}{0ex}}x+4y\ge 12\phantom{\rule{0ex}{0ex}}5x+8y=20\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 0 and y = 0
5x + 8y = 20 is already an equation.
Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.Clearly (0,0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x + y ≤ 8.
Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.
The line 5x + 8y = 20 is the line that passes through E(4, 0) and $F\left(0,\frac{5}{2}\right)$ .
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, x + y ≤ 8, x + 4y ≥ 12, 5x + 8y = 20, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are B(0,8), D(0,3), $G\left(\frac{20}{3},\frac{4}{3}\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = 30x + 20y 
B(0,8)  160 
D(0,3)  60 
$G\left(\frac{20}{3},\frac{4}{3}\right)$  266.66 
Therefore, the minimum value of Z is 60 at the point D(0,3). Hence, x = 0 and y =3 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 60.
Page No 30.32:
Question 13:
Maximize Z = 4x + 3y
Subject to
$3x+4y\le 24\phantom{\rule{0ex}{0ex}}8x+6y\le 48\phantom{\rule{0ex}{0ex}}x\le 5\phantom{\rule{0ex}{0ex}}y\le 6\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
We need to maximize Z = 4x + 3y
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5 , y = 6, x = 0 and y = 0.
The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0,6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24.So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0,8). Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation 8x + 6y ≤ 48. So, the region in xyplane that contains the origin represents the solution set of the given equation.
x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are O(0, 0), $G\left(5,0\right)$, $F\left(5,\frac{4}{3}\right)$, $E\left(\frac{24}{7},\frac{24}{7}\right)$ and $B\left(0,6\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = 4x + 3y 
O(0, 0)  4× 0 + 3 × 0 = 0 
$G\left(5,0\right)$  4 × 5 + 3 × 0 = 20 
$F\left(5,\frac{4}{3}\right)$  4 × 5 + 3 ×$\frac{4}{3}$ = 24 
$E\left(\frac{24}{7},\frac{24}{7}\right)$  4 × $\frac{24}{7}$ + 3 × $\frac{24}{7}$ = $\frac{196}{7}$= 24 
$B\left(0,6\right)$  4 × 0 + 3 × 6 = 18 
We see that the maximum value of the objective function Z is 24 which is at $F\left(5,\frac{4}{3}\right)$ and $E\left(\frac{24}{7},\frac{24}{7}\right)$.
Thus, the optimal value of Z is 24.
Page No 30.32:
Question 14:
Minimize Z = x − 5y + 20
Subject to
$xy\ge 0\phantom{\rule{0ex}{0ex}}x+2y\ge 2\phantom{\rule{0ex}{0ex}}x\ge 3\phantom{\rule{0ex}{0ex}}y\le 4\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 0, − x + 2y = 2, x = 3, y = 4, x = 0 and y = 0.
Region represented by x − y ≥ 0 or x ≥ y:
The line x − y = 0 or x = y passes through the origin.The region to the right of the line x = y will satisfy the given inequation.
Let's check by taking an example like if we take a point (4, 3) to the right of the line x = y .Here x ≥ y.So, it satisfy the given inequation. Take a point (4, 5) to the left of the line x = y. Here, x ≤ y. That means it does not satisfy the given inequation.
Region represented by − x + 2y ≥ 2:
The line − x + 2y = 2 meets the coordinate axes at A(−2, 0) and B(0, 1) respectively. By joining these points we obtain the line
− x + 2y = 2.Clearly (0,0) does not satisfies the inequation − x + 2y ≥ 2. So,the region in xy plane which does not contain the origin represents the solution set of the inequation − x + 2y ≥ 2 .
The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. x ≥ 3 is the region to the right of the line x = 3.
The line y = 4 is the line that passes through the point (0, 4) and is parallel to X axis. y ≤ 4 is the region below the line y = 4.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints x − y ≥ 0,− x + 2y ≥ 2, x ≥ 3, y ≤ 4, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are $C\left(3,\frac{5}{2}\right)$,$D\left(3,3\right)$, E(4, 4) and F(6, 4).
The values of Z at these corner points are as follows.
Corner point  Z = x − 5y + 20 
$C\left(3,\frac{5}{2}\right)$  3 − 5 × $\frac{5}{2}$ + 20 = $\frac{21}{2}$ 
$D\left(3,3\right)$  3 − 5 × 3 + 20 = 8 
E(4, 4)  4 − 5 × 4 + 20 = 4 
F(6, 4)  6 − 5 × 4 + 20 = 6 
Therefore, the minimum value of Z is 4 at the point E(4, 4). Hence, x = 4 and y = 4 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 4.
Page No 30.32:
Question 15:
Maximize Z = 3x + 5y
Subject to
$x+2y\le 20\phantom{\rule{0ex}{0ex}}x+y\le 15\phantom{\rule{0ex}{0ex}}y\le 5\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
We need to maximize Z = 3x + 5y
First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 20, x + y = 15, y = 5 , x = 0 and y = 0.
The line x + 2y = 20 meets the coordinate axis at A(20, 0) and B(0,10). Join these points to obtain the line x + 2y = 20.
Clearly, (0, 0) satisfies the inequation x + 2y ≤ 20. So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line x + y = 15 meets the coordinate axis at C(15, 0) and D(0,15). Join these points to obtain the line x + y = 15.
Clearly, (0, 0) satisfies the inequation x + y ≤ 15. So, the region in xyplane that contains the origin represents the solution set of the given equation.
y = 5 is the line passing through (0, 5) and parallel to the X axis.The region below the line y = 5 will satisfy the given inequation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are O(0, 0),$C\left(15,0\right)$, $E\left(10,5\right)$ and $F\left(0,5\right)$
The values of Z at these corner points are as follows.
Corner point  Z = 3x + 5y 
O(0, 0)  3 × 0 + 5 × 0 = 0 
$C\left(15,0\right)$  3 × 15 + 5 × 0 = 45 
$E\left(10,5\right)$  3 × 10 + 5 × 5 = 55 
$F\left(0,5\right)$  3 × 0 + 5 × 5 = 25 
We see that the maximum value of the objective function Z is 55 which is at $E\left(10,5\right)$.
Thus, the optimal value of Z is 55.
Page No 30.32:
Question 16:
Minimize Z = 3x_{1} + 5x_{2}
Subject to
${x}_{1}+3{x}_{2}\ge 3\phantom{\rule{0ex}{0ex}}{x}_{1}+{x}_{2}\ge 2\phantom{\rule{0ex}{0ex}}{x}_{1},{x}_{2}\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x_{1} + 3x_{2} = 3, x_{1} + x_{2} = 2, x_{1} = 0 and x_{2} = 0
Region represented by x_{1} + 3x_{2} ≥ 3 :
The line x_{1} + 3x_{2} = 3 meets the coordinate axes at A(3, 0) and B(0, 1) respectively. By joining these points we obtain the line x_{1} + 3x_{2} = 3.
Clearly (0,0) does not satisfies the inequation x_{1} + 3x_{2} ≥ 3 .So,the region in the plane which does not contain the origin represents the solution set of the inequation x_{1} + 3x_{2} ≥ 3.
Region represented by x_{1} + x_{2} ≥ 2:
The line x_{1} + x_{2} = 2 meets the coordinate axes at C(2, 0) and D(0, 2) respectively. By joining these points we obtain the line x_{1} + x_{2} = 2.Clearly (0,0) does not satisfies the inequation x_{1} + x_{2} ≥ 2. So,the region containing the origin represents the solution set of the inequation x_{1} + x_{2} ≥ 2.
Region represented by x_{1} ≥ 0 and x_{2} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x_{1} ≥ 0 and x_{2} ≥ 0.
The feasible region determined by the system of constraints, x_{1} + 3x_{2} ≥ 3 , x_{1} + x_{2} ≥ 2, x_{1} ≥ 0, and x_{2} ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), B(0, 1), $E\left(\frac{3}{2},\frac{1}{2}\right)$ and C(2, 0).
The values of Z at these corner points are as follows.
Corner point  Z = 3x_{1} + 5x_{2} 
O(0, 0)  3 × 0 + 5 × 0 = 0 
B(0, 1)  3 × 0 + 5 × 1 = 5 
$E\left(\frac{3}{2},\frac{1}{2}\right)$  3 × $\frac{3}{2}$ + 5 × $\frac{1}{2}$ = 7 
C(2, 0)  3 × 2 + 5 × 0 = 6 
Therefore, the minimum value of Z is 0 at the point O(0, 0). Hence, x_{1} = 0 and x_{2} = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 0.
Page No 30.33:
Question 17:
Maximize Z = 2x + 3y
Subject to
$x+y\ge 1\phantom{\rule{0ex}{0ex}}10x+y\ge 5\phantom{\rule{0ex}{0ex}}x+10y\ge 1\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 1, 10x +y = 5, x + 10y = 1, x = 0 and y = 0
Region represented by x + y ≥ 1:
The line x + y = 1 meets the coordinate axes at A(1, 0) and B(0,1) respectively. By joining these points we obtain the line x + y = 1.
Clearly (0,0) does not satisfies the inequation x + y ≥ 1. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 1.
Region represented by 10x +y ≥ 5:
The line 10x +y = 5 meets the coordinate axes at $C\left(\frac{1}{2},0\right)$ and D(0, 5) respectively. By joining these points we obtain the line
10x +y = 5.Clearly (0,0) does not satisfies the inequation 10x +y ≥ 5. So,the region which does not contains the origin represents the solution set of the inequation 10x +y ≥ 5.
Region represented by x + 10y ≥ 1:
The line x + 10y = 1 meets the coordinate axes at $A\left(1,0\right)$ and $F\left(0,\frac{1}{10}\right)$ respectively. By joining these points we obtain the line
x + 10y = 1.Clearly (0,0) does not satisfies the inequation x + 10y ≥ 1. So,the region which does not contains the origin represents the solution set of the inequation x + 10y ≥ 1.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≥ 1, 10x +y ≥ 5, x + 10y ≥ 1, x ≥ 0, and y ≥ 0, are as follows.
The feasible region is unbounded.Therefore, the maximum value is infinity i.e. the solution is unbounded.
Disclaimer:
The obtained answer is for the given question. Answer in the book is 2.It would be 2 if the question is to minimize Z instead of to maximize Z.
Page No 30.33:
Question 18:
Maximize Z = −x_{1} + 2x_{2}
Subject to
${x}_{1}+3{x}_{2}\le 10\phantom{\rule{0ex}{0ex}}{x}_{1}+{x}_{2}\le 6\phantom{\rule{0ex}{0ex}}{x}_{1}{x}_{2}\le 2\phantom{\rule{0ex}{0ex}}{x}_{1},{x}_{2}\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
−x_{1} + 3x_{2} = 10, x_{1} + x_{2} = 6, x_{1} + x_{2} = 2, x_{1} = 0 and x_{2} = 0
Region represented by −x_{1} + 3x_{2} ≤ 10:
The line −x_{1} + 3x_{2}_{ }= 10 meets the coordinate axes at A(−10, 0) and $B\left(0,\frac{10}{3}\right)$ respectively. By joining these points we obtain the line −x_{1} + 3x_{2}_{ }= 10.
Clearly (0,0) satisfies the inequation −x_{1} + 3x_{2} ≤ 10 .So,the region in the plane which contain the origin represents the solution set of the inequation
−x_{1} + 3x_{2} ≤ 10.
Region represented by x_{1} + x_{2} ≤ 6:
The line x_{1} + x_{2} = 6 meets the coordinate axes at C(6, 0) and D(0, 6) respectively. By joining these points we obtain the line x_{1} + x_{2} = 6.Clearly (0,0) satisfies the inequation x_{1} + x_{2} ≤ 6. So,the region containing the origin represents the solution set of the inequation x_{1} + x_{2} ≤ 6.
Region represented by x_{1}− x_{2} ≤ 2:
The line x_{1} − x_{2} = 2 meets the coordinate axes at E(2, 0) and F(0, −2) respectively. By joining these points we obtain the line x_{1} − x_{2} = 2.Clearly (0,0) satisfies the inequation x_{1}− x_{2} ≤ 2. So,the region containing the origin represents the solution set of the inequation x_{1}− x_{2} ≤ 2.
Region represented by x_{1} ≥ 0 and x_{2} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x_{1} ≥ 0 and x_{2} ≥ 0.
The feasible region determined by the system of constraints, −x_{1} + 3x_{2} ≤ 10, x_{1} + x_{2} ≤ 6, x_{1}− x_{2} ≤ 2, x_{1} ≥ 0, and x_{2} ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), E(2, 0), H(4, 2), G(2, 4) and $B\left(0,\frac{10}{3}\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = −x_{1} + 2x_{2} 
O(0, 0)  −1 × 0 + 2 × 0 = 0 
E(2, 0)  −1 × 2 + 2 × 0 = −2 
H(4, 2)  −1 × 4 + 2 × 2 = 0 
G(2, 4)  −1 × 2 + 2 × 4 = 6 
$B\left(0,\frac{10}{3}\right)$  −1 × 0 + 2 × $\frac{10}{3}$ = $\frac{20}{3}$ 
We see that the maximum value of the objective function Z is $\frac{20}{3}$ which is at $B\left(0,\frac{10}{3}\right)$.
Page No 30.33:
Question 19:
Maximize Z = x + y
Subject to
$2x+y\le 1\phantom{\rule{0ex}{0ex}}x\le 2\phantom{\rule{0ex}{0ex}}x+y\le 3\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
We need to maximize Z = x + y
First, we will convert the given inequations into equations, we obtain the following equations:
−2x + y = 1, x = 2, x + y = 3, x = 0 and y = 0.
The line −2x + y = 1 meets the coordinate axis at $A\left(\frac{1}{2},0\right)$ and B(0, 1). Join these points to obtain the line −2x + y = 1 .
Clearly, (0, 0) satisfies the inequation −2x + y ≤ 1. So, the region in xyplane that contains the origin represents the solution set of the given equation.
x = 2 is the line passing through (2, 0) and parallel to the Y axis.The region below the line x = 2 will satisfy the given inequation.
The line x + y = 3 meets the coordinate axis at C(3, 0) and D(0, 3). Join these points to obtain the line x + y = 3.
Clearly, (0, 0) satisfies the inequation x + y ≤ 3. So, the region in xyplane that contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are O(0, 0),$G\left(2,0\right)$, $E\left(2,1\right)$ and $F\left(\frac{2}{3},\frac{7}{3}\right)$
The values of Z at these corner points are as follows.
Corner point  Z = x + y 
O(0, 0)  0 + 0 = 0 
$C\left(2,0\right)$  2 + 0 = 2 
$E\left(2,1\right)$  2 +1 = 3 
$F\left(\frac{2}{3},\frac{7}{3}\right)$  $\frac{2}{3}+\frac{7}{3}=\frac{9}{3}=3$ 
We see that the maximum value of the objective function Z is 3 which is at $E\left(2,1\right)$ and $F\left(\frac{2}{3},\frac{7}{3}\right)$.
Thus, the optimal value of Z is 3.
Page No 30.33:
Question 20:
Maximize Z = 3x_{1} + 4x_{2}, if possible,
Subject to the constraints
${x}_{1}{x}_{2}\le 1\phantom{\rule{0ex}{0ex}}{x}_{1}+{x}_{2}\le 0\phantom{\rule{0ex}{0ex}}{x}_{1},{x}_{2}\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x_{1} − x_{2} = −1, −x_{1} + x_{2} = 0, x_{1} = 0 and x_{2} = 0
Region represented by x_{1} − x_{2} ≤ −1:
The line x_{1} − x_{2} = −1 meets the coordinate axes at A(−1, 0) and B(0, 1) respectively. By joining these points we obtain the line x_{1} − x_{2} = −1.
Clearly (0,0) does not satisfies the inequation x_{1} − x_{2} ≤ −1 .So,the region in the plane which does not contain the origin represents the solution set of the inequation x_{1} − x_{2} ≤ −1.
Region represented by −x_{1} + x_{2} ≤ 0 or x_{1}_{ }≥ x_{2}:
The line −x_{1} + x_{2} = 0 or x_{1}_{ }= x_{2}_{ }is the line passing through (0, 0).The region to the right of the line x_{1 }= x_{2} will satisfy the given inequation −x_{1} + x_{2} ≤ 0.
If we take a point (1, 3) to the left of the line x_{1 }= x_{2}. Here, 1≤3 which is not satifying the inequation x_{1}_{ }≥ x_{2}_{. }Therefore, region to the right of the line x_{1 }= x_{2} will satisfy the given inequation −x_{1} + x_{2} ≤ 0.
Region represented by x_{1} ≥ 0 and x_{2} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x_{1} ≥ 0 and x_{2} ≥ 0.
The feasible region determined by the system of constraints, x_{1} − x_{2} ≤ −1, −x_{1} + x_{2} ≤ 0, x_{1} ≥ 0, and x_{2} ≥ 0, are as follows.
We observe that the feasible region of the given LPP does not exist.
Page No 30.33:
Question 21:
Maximize Z = 3x + 3y, if possible,
Subject to the constraints
$xy\le 1\phantom{\rule{0ex}{0ex}}x+y\ge 3\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 1, x + y = 3, x = 0 and y = 0
Region represented by x − y ≤ 1:
The line x − y = 1 meets the coordinate axes at A(1, 0) and B(0, −1) respectively. By joining these points we obtain the line x − y = 1.
Clearly (0,0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x − y ≤ 1.
Region represented by x + y ≥ 3:
The line x + y = 3 meets the coordinate axes at C(3, 0) and D(0, 3) respectively. By joining these points we obtain the line x + y = 3.
Clearly (0,0) satisfies the inequation x + y ≥ 3. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 3.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints x − y ≤ 1, x + y ≥ 3, x ≥ 0 and y ≥ 0 are as follows.
The feasible region is unbounded.We would obtain the maximum value at infinity.
Therefore,maximum value will be infinity i.e. the solution is unboundedm
Page No 30.33:
Question 22:
Show the solution zone of the following inequalities on a graph paper:
$5x+y\ge 10\phantom{\rule{0ex}{0ex}}x+y\ge 6\phantom{\rule{0ex}{0ex}}x+4y\ge 12\phantom{\rule{0ex}{0ex}}x\ge 0,y\ge 0$
Find x and y for which 3x + 2y is minimum subject to these inequalities. Use a graphical method.
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
5x + y = 10, x +y = 6, x + 4y = 12, x = 0 and y = 0
Region represented by 5x + y ≥ 10:
The line 5x + y = 10 meets the coordinate axes at A(2, 0) and B(0, 10) respectively. By joining these points we obtain the line 5x + y = 10.
Clearly (0,0) does not satisfies the inequation 5x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 5x + y ≥ 10.
Region represented by x +y ≥ 6:
The line x +y = 6 meets the coordinate axes at C(6,0) and D(0, 6) respectively. By joining these points we obtain the line
2x +3y = 30.Clearly (0,0) does not satisfies the inequation x +y ≥ 6. So,the region which does not contain the origin represents the solution set of the inequation 2x +3y ≥ 30.
Region represented by x + 4y ≥ 12
The line x + 4y = 12 meets the coordinate axes at E(12, 0) and F(0, 3) respectively. By joining these points we obtain the line
x + 4y = 12.Clearly (0,0) does not satisfies the inequation x + 4y ≥ 12. So,the region which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + y ≥ 10, x +y ≥ 6,x + 4y ≥ 12, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are B(0, 10), G(1,5), H(4,2) and E(12,0).
The values of Z at these corner points are as follows.
Corner point  Z = 3x + 2y 
B(0, 10)  3 × 0 + 3 × 10 = 30 
G(1,5)  3 × 1 + 2 × 5 = 13 
H(4,2)  3 × 4 + 2 × 2 = 16 
E(12,0)  3 × 12 + 2 × 0 = 36 
Therefore, the minimum value of Z is 13 at the point G(1,5). Hence, x = 1 and y = 5 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 13.
Page No 30.33:
Question 23:
Find the maximum and minimum value of 2x + y subject to the constraints:
x + 3y ≥ 6, x − 3y ≤ 3, 3x + 4y ≤ 24, − 3x + 2y ≤ 6, 5x + y ≥ 5, x, y ≥ 0.
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.
The line x + 3y = 6 meets the coordinate axis at $A\left(6,0\right)$ and B(0, 2). Join these points to obtain the line x + 3y = 6.
Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 6. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line x − 3y = 3 meets the coordinate axis at C(3, 0) and D(0, −1). Join these points to obtain the line x − 3y = 3.
Clearly, (0, 0) satisfies the inequation x − 3y ≤ 3. So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line 3x + 4y = 24 meets the coordinate axis at E(8, 0) and F(0, 6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24. So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line −3x + 2y = 6 meets the coordinate axis at G(−2, 0) and H(0, 3). Join these points to obtain the line −3x + 2y = 6.
Clearly, (0, 0) satisfies the inequation −3x + 2y ≤ 6. So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line 5x + y = 5 meets the coordinate axis at $I\left(1,0\right)$ and J(0, 5). Join these points to obtain the line 5x + y = 5.
Clearly, (0, 0) does not satisfies the inequation 5x + y ≥ 5. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are P$\left(\frac{4}{13},\frac{45}{13}\right)$, K$\left(\frac{4}{3},5\right)$, L$\left(\frac{84}{13},\frac{15}{13}\right)$, M$\left(\frac{9}{2},\frac{1}{2}\right)$, N$\left(\frac{9}{14},\frac{25}{14}\right)$
The values of Z at these corner points are as follows.
Corner point  Z = 2x + y 
P$\left(\frac{4}{13},\frac{45}{13}\right)$  2 × $\frac{4}{13}$+ $\frac{45}{13}$ = $\frac{53}{13}$ 
K$\left(\frac{4}{3},5\right)$  2 × $\frac{4}{3}$+ 5 = $\frac{23}{3}$ 
L$\left(\frac{84}{13},\frac{15}{13}\right)$  2 × $\frac{84}{13}$+ $\frac{15}{13}$ = $\frac{183}{13}$ 
M$\left(\frac{9}{2},\frac{1}{2}\right)$  2 × $\frac{9}{2}$+ $\frac{1}{2}$ = $\frac{19}{2}$ 
N$\left(\frac{9}{14},\frac{25}{14}\right)$  2 × $\frac{9}{14}$+ $\frac{25}{14}$ = $\frac{43}{14}$ 
We see that the minimum value of the objective function Z is $\frac{43}{14}$ which is at N$\left(\frac{9}{14},\frac{25}{14}\right)$ and maximum value of the objective function is $\frac{183}{13}$ which is at L$\left(\frac{84}{13},\frac{15}{13}\right)$.
Page No 30.33:
Question 24:
Find the minimum value of 3x + 5y subject to the constraints
− 2x + y ≤ 4, x + y ≥ 3, x − 2y ≤ 2, x, y ≥ 0.
Answer:
First, we will convert the given inequations into equations, we obtain the following equations:
−2x + y = 4, x + y = 3, x − 2y = 2, x = 0 and y = 0.
The line −2x + y = 4 meets the coordinate axis at $A\left(2,0\right)$ and B(0, 4). Join these points to obtain the line −2x + y = 4.
Clearly, (0, 0) satisfies the inequation −2x + y ≤ 4. So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line x + y = 3 meets the coordinate axis at C(3, 0) and D(0, 3). Join these points to obtain the line x + y = 3.
Clearly, (0, 0) does not satisfies the inequation x + y ≥ 3. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line x − 2y = 2 meets the coordinate axis at E(2, 0) and F(0, −1). Join these points to obtain the line x − 2y = 2.
Clearly, (0, 0) satisfies the inequation x − 2y ≤ 2. So, the region in xyplane that contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are $B\left(0,4\right)$, $D\left(0,3\right)$ and $G\left(\frac{8}{3},\frac{1}{3}\right)$
The values of Z at these corner points are as follows.
Corner point  Z = 3x + 5y 
$B\left(0,4\right)$  3 × 0+ 5 × 4 = 20 
$D\left(0,3\right)$  3 × 0+ 5 × 3 = 15 
$G\left(\frac{8}{3},\frac{1}{3}\right)$  3 × $\frac{8}{3}$+ 5 × $\frac{1}{3}$ = $\frac{29}{3}$ 
We see that the minimum value of the objective function Z is $\frac{29}{3}$ which is at $G\left(\frac{8}{3},\frac{1}{3}\right)$.
Thus, the optimal value of Z is $\frac{29}{3}$.
Page No 30.33:
Question 25:
Solved the following linear programming problem graphically:
Maximize Z = 60x + 15y
Subject to constraints
$x+y\le 50\phantom{\rule{0ex}{0ex}}3x+y\le 90\phantom{\rule{0ex}{0ex}}x,y\ge 0$
Answer:
We have to maximize Z = 60x + 15y
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 50, 3x + y = 90, x = 0 and y = 0
Region represented by x + y ≤ 50:
The line x + y = 50 meets the coordinate axes at A(50,0) and B(0,50) respectively. By joining these points we obtain the line 3x + 5y = 15.
Clearly (0,0) satisfies the inequation x + y ≤ 50. So,the region containing the origin represents the solution set of the inequation x + y ≤ 50.
Region represented by 3x + y ≤ 90:
The line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively. By joining these points we obtain the line 3x + y = 90.
Clearly (0,0) satisfies the inequation 3x + y ≤ 90. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 90.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints, x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), C(30, 0), $E\left(20,30\right)$ and B(0, 50).
The values of Z at these corner points are as follows.
Corner point  Z = 60x + 15y 
O(0, 0)  60 × 0 + 15 × 0 = 0 
C(30, 0)  60 × 30 + 15 × 0 = 1800 
$E\left(20,30\right)$  60 × 20 + 15 × 30 =1650 
B(0, 50)  60 × 0 + 15 × 50 = 750 
Therefore, the maximum value of Z is $1800\mathrm{at}\mathrm{the}\mathrm{point}\left(30,0\right)$.Hence, x = 30 and y = 0 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 1800.
Page No 30.33:
Question 26:
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0 [CBSE 2015]
Answer:
The given linear programming problem is Maximize Z = 2x + 5y
subject to the constraints
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0
Converting the inequations into equations, we obtain the following equations of straight lines:
2x + 4y = 8, 3x + y = 6, x + y = 4
The line 2x + 4y = 8 meets the coordinate axes at (4, 0) and (0, 2).
The line 3x + y = 6 meets the coordinate axes at (2, 0) and (0, 6).
The line x + y = 4 meets the coordinate axes at (4, 0) and (0, 4).
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(0, 2), B$\left(\frac{8}{5},\frac{6}{5}\right)$ and C(2, 0).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 2x + 5y 
(0, 0)  2 × 0 + 5 × 0 = 0 
(2, 0)  2 × 2 + 5 × 0 = 4 
(0, 2)  2 × 0 + 5 × 2 = 10 → Maximum 
$\left(\frac{8}{5},\frac{6}{5}\right)$  $2\times \frac{8}{5}+5\times \frac{6}{5}=\frac{46}{5}$ 
Thus, the maximum value of Z is 10 at x = 0, y = 2.
Page No 30.33:
Question 27:
Solve the following LPP graphically:
Maximize Z$=$20$x+$ 10$y$
Subject to the following constraints
$x+$2$y\le $28
3$x+$$y\le $24
$x\ge $ 2
$x$,$y\ge $ 0
Answer:
The given constraints are
x + 2y ≤ 28
3x + y ≤ 24
x ≥ 2
x, y ≥ 0
Converting the given inequations into equations, we get
x + 2y = 28, 3x + y = 24, x = 2, x = 0 and y = 0
These lines are drawn on the graph and the shaded region ABCD represents the feasible region of the given LPP.
It can be observed that the feasible region is bounded. The coordinates of the corner points of the feasible region are A(2, 13), B(2, 0), C(4, 12) and D(8, 0).
The values of the objective function, Z at these corner points are given in the following table:
Corner Point  Value of the Objective Function Z = 20x + 10y 
A(2, 13)  Z = 20 × 2 + 10 × 13 = 170 
B(2, 0)  Z = 20 × 2 + 10 × 0 = 40 
C(8, 0)  Z = 20 × 8 + 10 × 0 = 160 
D(4, 12)  Z = 20 × 4 + 10 × 12 = 200 
From the table, Z is maximum at x = 4 and y = 12 and the maximum value of Z is 200.
Thus, the maximum value of Z is 200.
Page No 30.33:
Question 28:
Solve the following linear programming problem graphically:
Minimize $z=$6$x+$3$y$
Subject to the constraints:
4$x+$$y\ge $80
$x+$5$y\ge $115
3$x+$2$y\le $150
$x\ge $0, $y\ge $0
Answer:
The given constraints are
4x + y ≥ 80
x + 5y ≥ 115
3x + 2y ≤ 150
x, y ≥ 0
Converting the given inequations into equations, we get
4x + y = 80, x + 5y = 115, 3x + 2y = 150, x = 0 and y = 0
These lines are drawn on the graph and the shaded region ABC represents the feasible region of the given LPP.
It can be observed that the feasible region is bounded. The coordinates of the corner points of the feasible region are A(2, 72), B(15, 20) and C(40, 15).
The values of the objective function, Z at these corner points are given in the following table:
Corner Point  Value of the Objective Function Z = 6x + 3y 
A(2, 72)  Z = 6 × 2 + 3 × 72 = 228 
B(15, 20)  Z = 6 × 15 + 3 × 20 = 150 
C(40, 15)  Z = 6 × 40 + 3 × 15 = 285 
From the table, Z is minimum at x = 15 and y = 20 and the minimum value of Z is 150.
Thus, the minimum value of Z is 150.
Page No 30.38:
Question 1:
A diet of two foods F_{1} and F_{2} contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per 25 gms) is given in the following table:
Nutrients 
Food 
F_{1}  F_{2} 
Thiamine  0.25  0.10 

Phosphorous  0.75  1.50  
Iron  1.60  0.80 
Answer:
Let 25x grams of food F_{1}_{ }and 25y grams of food F_{2} be used to fulfil the minimum requirement of thiamine, phosphorus and iron.
As, we are given
Nutrients 
Food 
F_{1}  F_{2} 
Thiamine  0.25  0.10  
Phosphorous  0.75  1.50  
Iron  1.60  0.80 
And the minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron.
$\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}0.25x+0.10y\ge 1\phantom{\rule{0ex}{0ex}}0.75x+1.50y\ge 7.5\phantom{\rule{0ex}{0ex}}1.6x+0.8y\ge 10\phantom{\rule{0ex}{0ex}}\mathrm{Since},\mathrm{the}\mathrm{quantity}\mathrm{cannot}\mathrm{be}\mathrm{negative}\phantom{\rule{0ex}{0ex}}\therefore x,y\ge 0$
The cost of F_{1} is 20 paise per 25 gms while the cost of F_{2} is 15 paise per 25 gms.Therefore, the cost of 25x grams of food F_{1} and 25y grams of food F_{2} is
Rs (0.20x + 0.15y).
Hence,
Minimize Z = $0.20x+0.15y$
subject to
$0.25x+0.10y\ge 1\phantom{\rule{0ex}{0ex}}0.75x+1.50y\ge 7.5\phantom{\rule{0ex}{0ex}}1.6x+0.8y\ge 10\phantom{\rule{0ex}{0ex}}x,y\ge 0$
First, we will convert the given inequations into equations, we obtain the following equations:
0.25x + 0.10y = 1, 0.75x + 1.50y = 7.5, 1.6x + 0.8y = 10, x = 0 and y = 0.
The line 0.25x + 0.10y = 1 meets the coordinate axis at $A\left(4,0\right)$ and B(0, 10). Join these points to obtain the line 0.25x + 0.10y = 1.
Clearly, (0, 0) does not satisfies the inequation 0.25x + 0.10y ≥ 1. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line 0.75x + 1.50y = 7.5. meets the coordinate axis at C(10, 0) and D(0, 5). Join these points to obtain the line 0.75x + 1.50y = 7.5.
Clearly, (0, 0) does not satisfies the inequation 0.75x + 1.50y ≥ 0.75. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line 1.6x + 0.8y = 10 meets the coordinate axis at $E\left(\frac{25}{4},0\right)$ and $F\left(0,\frac{25}{2}\right)$. Join these points to obtain the line 1.6x + 0.8y = 10.
Clearly, (0, 0) does not satisfies the inequation 1.6x + 0.8y ≥ 10. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The corner points of the feasible region are F(0, 12.5), G(5, 2.5), C(10, 0)
The value of the objective function at these points are given by the following table
Points  Value of Z 
F  0.20(0)+0.15(12.5) = 1.875 
G  0.20(5)+0.15(2.5) = 1.375 
C  0.20(10) + 0.15(0) = 200 
Thus, the minimum cost is at G which is Rs 1.375
Page No 30.38:
Question 2:
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?
Answer:
Let the sick person takes x units and y units of food I and II respectively that were taken in the diet.
Since, per unit of food I costs Rs 4 and that of food II costs Rs 3.
Therefore, x units of food I costs Rs 4x and y units of food II costs Rs 3y.
Total cost = Rs (4x + 3y)
Let Z denote the total cost
Then, Z = 4x + 3y
If one unit of A contains 200 units of vitamin and one unit of food B contains 100 units of vitamin.
Thus, x units of food I and y units of food II contains 200x + 100y units of vitamin.
But a diet for a sick person must contain at least 4000 units of vitamins.
$\therefore $ $200x+100y\ge 4000$
If one unit of A contains 1 unit of mineral and one unit of food B contains 2 units of mineral.
Thus, x units of food I and y units of food II contains x + 2y units of mineral.
But a diet for a sick person must contain at least 50 units of vitamins.
$\therefore $ $x+2y\ge 50$
If one unit of A contains 40 calories and one unit of food B contains 40 calories.
Thus, x units of food I and y units of food II contains 40x + 40y units of calories.
But a diet for a sick person must contain at least 1400 calories.
$\therefore 40x+40y\ge 1400$
Finally, the quantities of food I and food II are non negative values.
So, $x,y\ge 0$
Hence, the required LPP is as follows:
Min Z = 4x + 3y
subject to
$200x+100y\ge 4000\phantom{\rule{0ex}{0ex}}x+2y\ge 50\phantom{\rule{0ex}{0ex}}40x+40y\ge 1400\phantom{\rule{0ex}{0ex}}x,y\ge 0$
First, we will convert the given inequations into equations, we obtain the following equations:
200x + 100y = 4000, x +2y = 50, 40x + 40y =1400, x = 0 and y = 0
Region represented by 200x + 100y ≥ 4000:
The line 200x + 100y = 4000 meets the coordinate axes at A_{1}(20, 0) and B_{1}(0,40) respectively. By joining these points we obtain the line
200x + 100y = 4000.Clearly (0,0) does not satisfies the inequation 200x + 100y ≥ 4000. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 200x + 100y ≥ 4000.
Region represented by x +2y ≥ 50:
The line x +2y = 50 meets the coordinate axes at C_{1}(50, 0) and D_{1}(0, 25) respectively. By joining these points we obtain the line
x +2y = 50.Clearly (0,0) does not satisfies the x +2y ≥ 50. So,the region which does not contains the origin represents the solution set of the inequation x +2y ≥ 50.
Region represented by 40x + 40y ≥ 1400:
The line 40x + 40y = 1400 meets the coordinate axes at E_{1}(35, 0) and F_{1}(0, 35) respectively. By joining these points we obtain the line
40x + 40y = 1400.Clearly (0,0) does not satisfies the inequation 40x + 40y ≥ 1400. So,the region which does not contains the origin represents the solution set of the inequation 40x + 40y ≥ 1400.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 200x + 100y ≥ 4000,x +2y ≥ 50, 40x + 40y ≥ 1400, x ≥ 0, and y ≥ 0 are as follows.
The corner points of the feasible region are B_{1}(0, 40), G_{1}(5, 30), H_{1}(20, 15) and C_{1}(50, 0)
The value of the objective function at these points are given by the following table
Points  Value of Z 
B_{1}  4(0)+3(40) = 120 
G_{1}  4(5)+3(30) = 110 
H_{1}  4(20) + 3(15) = 125 
C_{1}  4(50)+3(0) = 200 
The minimum cost is Rs 110 which is at G_{1}(5, 30).
Hence, the required combination of food is 5 units of food A and 30 units of food B.
Page No 30.38:
Question 3:
To maintain one's health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:
Food I  Food II  Minimum daily requirement  
Calcium Protein Calories 
10 5 2 
4 6 6 
20 20 12 
Price  Rs 0.60 per unit  Rs 1.00 per unit 
Answer:
Let the person takes x units and y units of food I and II respectively that were taken in the diet.
Since, per unit of food I costs Re 0.60 and that of food II costs Re 1.00.
Therefore, x lbs of food I costs Rs 0.60x and y lbs of food II costs Rs 1.00y.
Total cost per day = Rs (0.60x +1.00y)
Let Z denote the total cost per day
Then, Z = 0.60x +1.00y
Since, each unit of food I contains 10 units of calcium.Therefore, x units of food I contains 10x units of calcium.
Each unit of food II contains 4 units of calcium.So, y units of food II contains 4y units of calcium.
Thus, x units of food I and y units of food II contains (10x + 4y) units of calcium.
But, the minimum requirement is 20 units of calcium.
$\therefore $ $10x+4y\ge 20$
Since, each unit of food I contains 5 units of protein.Therefore, x units of food I contains 5x units of protein.
Each unit of food II contains 6 units of protein.So,y units of food II contains 6y units of protein.
Thus, x units of food I and y units of food II contains (5x + 6y) units of protein.
But, the minimum requirement is 20 lbs of protein.
$\therefore $ $5x+6y\ge 20$
Since, each unit of food I contains 2 units of calories.Therefore, x units of food I contains 2x units of calories.
Each unit of food II contains 6 units of calories.So,y units of food II contains 6y units of calories.
Thus, x units of food I and y units of food II contains (2x + 6y) units of calories.
But, the minimum requirement is 12 lbs of calories.
$\therefore 2x+6y\ge 12$
Finally, the quantities of food I and food II are non negative values.
So, $x,y\ge 0$
Hence, the required LPP is as follows:
Min Z = 0.60x + 1.00y
subject to
$10x+4y\ge 20\phantom{\rule{0ex}{0ex}}5x+6y\ge 20\phantom{\rule{0ex}{0ex}}2x+6y\ge 12\phantom{\rule{0ex}{0ex}}x,y\ge 0$
First, we will convert the given inequations into equations, we obtain the following equations:
10x + 4y = 20, 5x +6y = 20, 2x + 6y =12, x = 0 and y = 0
Region represented by 10x + 4y ≥ 20:
The line 10x + 4y = 20 meets the coordinate axes at A(2, 0) and B(0, 5) respectively. By joining these points we obtain the line
10x + 4y = 20.Clearly (0,0) does not satisfies the inequation 10x + 4y ≥ 20. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 10x + 4y ≥ 20.
Region represented by 5x +6y ≥ 20:
The line 5x +6y = 20 meets the coordinate axes at $C\left(4,0\right)$ and $D\left(0,\frac{10}{3}\right)$ respectively. By joining these points we obtain the line
5x +6y = 20.Clearly (0,0) does not satisfies the 5x +6y ≥ 20. So,the region which does not contains the origin represents the solution set of the inequation 5x +6y ≥ 20.
Region represented by 2x + 6y ≥ 12:
The line 2x + 6y =12 meets the coordinate axes at $E\left(6,0\right)$ and $F\left(0,2\right)$ respectively. By joining these points we obtain the line
2x + 6y =12.Clearly (0,0) does not satisfies the inequation 2x + 6y ≥ 12. So,the region which does not contains the origin represents the solution set of the inequation 2x + 6y ≥ 12.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 10x + 4y ≥ 20, 5x +6y ≥ 20, 2x + 6y ≥ 12, x ≥ 0, and y ≥ 0 are as follows.
The set of all feasible solutions of the above LPP is represented by the feasible region shaded in the graph.
The corner points of the feasible region are B(0, 5), G$\left(1,\frac{5}{2}\right)$, H$\left(\frac{8}{3},\frac{10}{9}\right)$ and E$\left(6,0\right)$
The value of the objective function at these points are given by the following table
Points  Value of Z 
B  $0.6\left(0\right)+5=5$ 
G  $0.6\left(1\right)+\frac{5}{2}=3.1$ 
H  $0.6\left(\frac{8}{3}\right)+\left(\frac{10}{9}\right)=1.6+1.1=2.7$ 
E  $0.6\left(6\right)+\left(0\right)=3.6$ 
We see that the minimum cost is 2.7 which is at $\left(\frac{8}{3},\frac{10}{9}\right)$.
Thus, at minimum cost, $\frac{8}{3}$ units of food I and $\frac{10}{9}$ units of food II should be included in the diet.
Page No 30.39:
Question 4:
A hospital dietician wishes to find the cheapest combination of two foods, A and B, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of A contains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?
Answer:
Let the dietician wishes to mix x units of food A and y units of food B.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows
Thiamine(mg)  Calories  
Food A  0.12  100 
Food B  0.1  150 
Minimum requirement  0.5  600 
According to the question,
The constraints are
$0.12x+0.1y\ge 0.5\phantom{\rule{0ex}{0ex}}100x+150y\ge 600$
It is given that each food costs 10 paise per units
Therefore,
Total cost, Z = $10x+10y$
Thus, the mathematical formulation of the given linear programmimg problem is
$0.12x+0.1y\ge 0.5\phantom{\rule{0ex}{0ex}}100x+150y\ge 600$
Region represented by 0.12x +0.1y ≥ 0.5:
The line 0.12x + 0.6y = 20 meets the coordinate axes at ${A}_{1}\left(\frac{25}{6},0\right)$ and ${B}_{1}\left(0,5\right)$ respectively. By joining these points we obtain the line
0.12x + 0.6y = 20.Clearly (0,0) does not satisfies the 0.12x + 0.6y = 20. So,the region which does not contains the origin represents the solution set of the inequation 0.12x +0.1y ≥ 0.5.
Region represented by 100x + 150y ≥ 600:
The line 100x + 150y = 600 meets the coordinate axes at ${C}_{1}\left(6,0\right)$ and ${D}_{1}\left(0,4\right)$ respectively. By joining these points we obtain the line 100x + 150y = 600. Clearly (0,0) does not satisfies the inequation 100x + 150y ≥ 600. So,the region which does not contains the origin represents the solution set of the inequation 100x + 150y ≥ 600.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 0.12x +0.1y ≥ 0.5, 100x + 150y ≥ 600, x ≥ 0, and y ≥ 0 are as follows.
The corner points are B_{1}(0, 5), ${E}_{1}\left(\frac{15}{8},\frac{11}{4}\right)$, ${C}_{1}\left(6,0\right)$
The values of Z at these corner points are as follows
Corner point  Z= 10x +10y 
B_{1}  50 
E_{1}  46.2 
C_{1}  60 
The minimum value of Z is at ${E}_{1}\left(\frac{15}{8},\frac{11}{4}\right)$.
Hence, cheapest combination of foods will be 1.875 units of food A and 2.75 units of food B.
Page No 30.39:
Question 5:
A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below:
Vitamin A 
Vitamin B 
Vitamin 
Vitamin D 

Food X Food Y 
1 2 
1 1 
1 3 
2 1 
Answer:
Let the dietician wishes to mix x kg of food X and y kg of food Y.
Therefore, $x,y\ge 0$
As we are given,
Vitamin A  Vitamin B 
Vitamin C 
Vitamin D  
Food X Food Y 
1 2 
1 1 
1 3 
2 1 
It is given that the mixture should contain at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D.
Therefore, the constraints are
$x+2y\ge 6\phantom{\rule{0ex}{0ex}}x+y\ge 7\phantom{\rule{0ex}{0ex}}x+3y\ge 11\phantom{\rule{0ex}{0ex}}2x+y\ge 9$
It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs 8 per kg.
Thus, Z = $5x+8y$
Thus, the mathematical formulation of the given linear programmimg problem is
Minimize Z = $5x+8y$
subject to
$x+2y\ge 6\phantom{\rule{0ex}{0ex}}x+y\ge 7\phantom{\rule{0ex}{0ex}}x+3y\ge 11\phantom{\rule{0ex}{0ex}}2x+y\ge 9$
First, we will convert the given inequations into equations, we obtain the following equations:
x + 2y = 6, x + y = 7, x + 3y =11, 2x + y =9, x = 0 and y = 0.
The line x + 2y = 6 meets the coordinate axis at A_{1}(6, 0) and B_{1}(0, 3). Join these points to obtain the line x + 2y = 6. Clearly, (0, 0) does not satisfies the inequation x + 2y ≥ 6. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line x + y = 7 meets the coordinate axis at C_{1}(7, 0) and D_{1}(0, 7). Join these points to obtain the line x + y = 7. Clearly, (0, 0) does not satisfies the inequation x + y ≥ 7. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line x + 3y = 11 meets the coordinate axis at ${E}_{1}\left(11,0\right)$ and ${F}_{1}\left(0,\frac{11}{3}\right)$. Join these points to obtain the line x + 3y = 11.Clearly, (0, 0) does not satisfies the inequation x + 3y ≥ 11. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line 2x + y = 9 meets the coordinate axis at ${G}_{1}\left(\frac{9}{2},0\right)$ and ${H}_{1}\left(0,9\right)$. Join these points to obtain the line 2x + y = 9.Clearly, (0, 0) does not satisfies the inequation 2x + y ≥ 9. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The feasible region determined by the system of constraints is
The corner points are H_{1}(0, 9), I_{1}(2 ,5), J_{1}(5, 2), E_{1}(11, 0).
The values of Z at these corner points are as follows
Corner point  Z= 5x + 8y 
H_{1}  72 
I_{1}  50 
J_{1}  41 
E_{1}  55 
The minimum value of Z is at J_{1}(5, 2) which is Rs 41.
Hence, cheapest combination of foods will be 5 units of food X and 2 units of food Y.
Page No 30.39:
Question 6:
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F_{1} and F_{2} are available. Food F_{1} costs Rs 4 per unit and F_{2} costs Rs 6 per unit one unit of food F_{1} contains 3 units of vitamin A and 4 units of minerals. One unit of food F_{2} contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements.
Answer:
Let the dietician wishes to mix x units of food F_{1} and y units of food F_{2}.
Clearly, $x,y\ge 0$
The given information can be tabulated as follows:
Vitamin A  Minerals  
Food F_{1}  3  4 
Food F_{2}  6  3 
Minimum requirement  80  100 
The constraints are
$3x+6y\ge 80\phantom{\rule{0ex}{0ex}}4x+3y\ge 100$
It is given that cost of food F_{1}_{ }and F_{2} is Rs 4 and Rs 6 per unit respectively. Therefore, cost of x units of food F_{1} and y units of food F_{2} is Rs 4x and Rs 6y respectively.
Let Z denote the total cost
∴ Z = 4x + 6y
Thus, the mathematical formulation of the given linear programmimg problem is
Minimize $\mathrm{Z}=4x+6y$ subject to
$3x+6y\ge 80\phantom{\rule{0ex}{0ex}}4x+3y\ge 100$
$x,y\ge 0$
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 6y = 80, 4x + 3y = 100, x = 0 and y = 0
The line 3x + 6y = 80 meets the coordinate axis at $A\left(\frac{80}{3},0\right)$ and $B\left(0,\frac{40}{3}\right)$. Join these points to obtain the line 3x + 6y = 80. Clearly, (0, 0) does not satisfies the inequation 3x + 6y ≥ 80. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line 4x + 3y = 100 meets the coordinate axis at C(25, 0) and $D\left(0,\frac{100}{3}\right)$. Join these points to obtain the line 4x + 3y = 100. Clearly, (0, 0) does not satisfies the inequation 4x + 3y ≥ 100. So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The feasible region determined by the system of constraints is
The corner points are D$\left(0,\frac{100}{3}\right)$, E$\left(24,\frac{4}{3}\right)$ and $A\left(\frac{80}{3},0\right)$.
The values of Z at these corner points are as follows:
Corner point  Z= 4x + 6y 
D$\left(0,\frac{100}{3}\right)$  200 
E$\left(24,\frac{4}{3}\right)$  104 
$A\left(\frac{80}{3},0\right)$  $\frac{320}{3}$ 
The minimum value of Z is Rs 104 which is at E$\left(24,\frac{4}{3}\right)$.
Page No 30.39:
Question 7:
Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs 5 per kg and rice costs Rs 4 per kg.
Answer:
Let the cereal contain x kg of bran and y kg of rice.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows
Protein(grams)  Iron(milligrams)  
Bran  80  40 
Rice  100  30 
minimum requirement  88  36 
Bran and rice contains at least 88 grams of protein and at least 36 milligrams of iron.
Thus,the constraints are
$80x+100y\ge 88\phantom{\rule{0ex}{0ex}}40x+30y\ge 36\phantom{\rule{0ex}{0ex}}$
It is given that bran costs Rs 5 per kg and rice costs Rs 4 per kg. Therefore, cost of x kg of bran and y kg of rice is Rs 5x and Rs 4y respectively.
Hence, total profit is Rs (5x + 4y)
Let Z denote the total cost.
$\therefore $ Z = $5x+4y$
Thus, the mathematical formulation of the given problem is
Minimize $Z=5x+4y$
subject to
$80x+100y\ge 88\phantom{\rule{0ex}{0ex}}40x+30y\ge 36\phantom{\rule{0ex}{0ex}}$
$x,y\ge 0$
First we will convert inequations into equations as follows :
80x + 100y = 60, 40x + 30y = 36, x = 0 and y = 0
Region represented by 80x + 100y ≥ 88:
The line 80x + 100y = 60 meets the coordinate axes at A(1.1, 0) and B(0, 0.88)respectively. By joining these points we obtain the line 80x + 100y = 60. Clearly (0,0) does not satisfies the 80x + 100y ≥ 88. So,the region which does not contains the origin represents the solution set of the inequation
80x + 100y ≥ 88.
Region represented by 40x + 30y ≥ 36:
The line 40x + 30y = 36 meets the coordinate axes at C(0.9, 0) and D(0, 1.2) respectively. By joining these points we obtain the line 40x + 30y = 36.
Clearly (0,0) does not satisfies the inequation 40x + 30y ≥ 36. So,the region which does not contains the origin represents the solution set of the inequation 40x + 30y ≥ 36.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 80x + 100y ≥ 88, 40x + 30y ≥ 36, x ≥ 0, and y ≥ 0 are as follows.
The feasible region determined by the system of constraints is
The corner points are D(0, 1.2), E(0.6, 0.4) and A(1.1, 0).
The values of Z at these corner points are as follows:
Corner point  Z = 5x + 4y 
D(0, 1.2)  4.8 
E(0.6, 0.4)  4.6 
A(1.1, 0)  5.5 
The minimum value of Z is 4.6 which is attained at E(0.6, 0.4).
Hence, the minimum cost is Rs 4.6.
Page No 30.39:
Question 8:
A wholesale dealer deals in two kinds, A and B (say) of mixture of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew nuts and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew nuts and 180 grams of hazel nuts. The remainder of both mixtures is per nuts. The dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew nuts and 540 grams of hazel nuts. Mixture A costs Rs 8 per kg. and mixture B costs Rs 12 per kg. Assuming that mixtures A and B are uniform, use graphical method to determine the number of kg. of each mixture which he should use to minimise the cost of the bag.
Answer:
Let x kg of kind A and y kg of kind B were used.
Quantity cannot be negative.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Nut  Almonds(grams)  Cashewnuts(grams)  Hazel nuts(grams) 
A(x)  60  30  30 
B(y)  30  60  180 
Availability  240  300  540 
Therefore, the constraints are
$60x+30y\ge 240\phantom{\rule{0ex}{0ex}}30x+60y\ge 300\phantom{\rule{0ex}{0ex}}30x+180y\ge 540$
Mixture A costs Rs 8 per kg. and mixture B costs Rs 12 per kg.
Total cost = Z = $8x+12y$ which is to be minimised.
Thus, the mathematical formulation of the given linear programmimg problem is
Min Z = $8x+12y$
subject to
$2x+y\ge 8\phantom{\rule{0ex}{0ex}}x+2y\ge 10\phantom{\rule{0ex}{0ex}}x+6y\ge 18$
$x,y\ge 0$
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 8, x +2y = 10, x +6y = 18, x = 0 and y = 0
Region represented by 2x + y ≥ 8:
The line 2x + y = 8 meets the coordinate axes at A_{1}(4, 0) and B_{1}(0, 8) respectively. By joining these points we obtain the line 2x + y = 8.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 8. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 8.
Region represented by x +2y ≥ 10:
The line x +2y = 10 meets the coordinate axes at C_{1}(10,0) and D_{1}(0, 5) respectively. By joining these points we obtain the line
x +2y = 10. Clearly (0,0) does not satisfies the inequation x +2y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x +2y ≥ 10.
Region represented by x +6y ≥ 18:
The line x +6y = 18 meets the coordinate axes at E_{1}(18,0) and F_{1}(0, 3) respectively. By joining these points we obtain the line x + 6y = 18.Clearly (0,0) does not satisfies the inequation x + 6y ≥ 18. So,the region which does not contain the origin represents the solution set of the inequation x +6y ≥ 18.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + y ≥ 8, x + 2y ≥ 10,x + 6y ≥ 18, x ≥ 0, and y ≥ 0, are as follows.
The corner points are B_{1}(0, 8), G_{1}(2, 4), H_{1}(6, 2) and E_{1}(18, 0).
The values of Z at these corner points are as follows
Corner point  Z= 8x + 12y 
B_{1}  96 
G_{1}  64 
H_{1}  72 
E_{1}  144 
The minimum value of Z is 64 which is attained at G_{1}$\left(2,4\right)$.
Thus, the minimum cost is Rs 64 obtained when 2 units of kind A and 4 units of kind B nuts were used.
Page No 30.39:
Question 9:
One kind of cake requires 300 gm of flour and 15 gm of fat, another kind of cake requires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 gm of fat, assuming that there is no shortage of the other ingradients used in making the cake. Make it as an LPP and solve it graphically.
Answer:
Let there be x cakes of first kind and y cakes of second kind.
∴ x ≥ 0 and y ≥ 0
The given information can be compiled in the form of a table as:
Flour (gm)  Fat (gm)  
Cakes of first kind  300  15 
Cakes of second kind  150  30 
Availability  7500  600 
Therefore, constraints are:
$300x+150y\le 7500\phantom{\rule{0ex}{0ex}}15x+30y\le 600$
Total number of cakes that can be made = Z = x + y
Thus, the given problem can be mathematically formulated as:
Maximize Z = x + y
subject to constraints,
$300x+150y\le 7500\phantom{\rule{0ex}{0ex}}15x+30y\le 600$
x, y ≥ 0
First we will convert inequations into equations as follows:
300x + 150y = 7500, 15x + 30y = 600, x = 0 and y = 0
Region represented by 300x + 150y ≤ 7500:
The line 300x + 150y = 7500 meets the coordinate axes at A(25, 0) and $B\left(0,50\right)$ respectively. By joining these points we obtain the line
300x + 150y = 7500. Clearly (0,0) satisfies the 300x + 150y = 7500. So, the region which contains the origin represents the solution set of the inequation 300x + 150y ≤ 7500.
Region represented by 15x + 30y ≤ 600:
The line 15x + 30y = 600 meets the coordinate axes at C(40, 0) and $D\left(0,20\right)$ respectively. By joining these points we obtain the line
15x + 30y = 600. Clearly (0,0) satisfies the inequation 15x + 30y ≤ 600. So,the region which contains the origin represents the solution set of the inequation 15x + 30y ≤ 600.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 300x + 150y ≤ 7500, 15x + 30y ≤ 600, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), A(25, 0), E(20, 10) and D(0, 20).
The values of Z at these corner points are:
Corner points 
Z = x + y 
O(0, 0)  0 
A(25, 0)  25 
E(20, 10)  30 
D(0, 20)  20 
Thus, maximum numbers of cakes that can be made from 20 of the first kind and 10 of the other kind.
Page No 30.40:
Question 10:
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹60/kg and food Q costs ₹80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Answer:
Let x units of food P and y units of food Q are mixed together to make the mixture.
The cost of food P is ₹60/kg and that of Q is ₹80/kg. So, x kg of food P and y kg of food Q will cost ₹(60x + 80y).
Since one kg of food P contains 3 units of vitamin A and one kg of food Q contains 4 units of vitamin A, therefore, x kg of food P and y kg of food Q will contain (3x + 4y) units of vitamin A. But, the mixture should contain atleast 8 units of vitamin A.
∴ 3x + 4y ≥ 8
Similarly, x kg of food P and y kg of food Q will contain (5x + 2y) units of vitamin B. But, the mixture should contain atleast 11 units of vitamin B.
∴ 5x + 2y ≥ 11
Thus, the given linear programming problem is
Minimise Z = 60x + 80y
subject to the constraints
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A$\left(\frac{8}{3},0\right)$, B$\left(2,\frac{1}{2}\right)$ and C$\left(0,\frac{11}{2}\right)$.
The value of the objective function at these points are given in the following table.
Corner Point  Z = 60x + 80y 
$\left(\frac{8}{3},0\right)$  $60\times \frac{8}{3}+80\times 0=160$ → Minimum 
$\left(2,\frac{1}{2}\right)$  $60\times 2+80\times \frac{1}{2}=160$ → Minimum 
$\left(0,\frac{11}{2}\right)$  $60\times 0+80\times \frac{11}{2}=440$ 
The smallest value of Z is 160 which is obtained at the points $\left(\frac{8}{3},0\right)$ and $\left(2,\frac{1}{2}\right)$.
It can be verified that the open halfplane represented by 60x + 80y < 160 has no common points with the feasible region.
So, the minimum value of Z is 160. Hence, the minimum cost of the mixture is ₹160.
Page No 30.40:
Question 11:
One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no storage of the other ingredients used in making the cakes.
Answer:
Let the number of cakes of one kind and another kind be x and y, respectively.
Therefore, the total number of cakes produced are (x + y).
One kind of cake requires 200 g of flour and another kind of cake requires 100 g of flour. So, x cakes of one kind and y cakes of another kind requires (200x + 100y) g of flour. But, cakes should contain atmost 5 kg of flour.
∴ 200x + 100y ≤ 5000
⇒ 2x + y ≤ 50
One kind of cake requires 25 g of fat and another kind of cake requires 50 g of fat. So, x cakes of one kind and y cakes of another kind requires (25x + 50y) g of fat. But, cakes should contain atmost 1 kg of fat.
∴ 25x + 50y ≤ 1000
⇒ x + 2y ≤ 40
Thus, the given linear programming problem is
Minimise Z = x + y
subject to the constraints
2x + y ≤ 50
x + 2y ≤ 40
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(25, 0), B(20, 10) and C(0, 20).
The value of the objective function at these points are given in the following table.
Corner Point  Z = x + y 
(0, 0)  0 + 0 = 0 
(25, 0)  25 + 0 = 25 
(20, 10)  20 + 10 = 30 → Maximum 
(0, 20)  0 + 20 = 20 
Thus, the maximum value of Z is 30 at x = 20, y = 10.
Hence, the maximum number of cakes which can be made are 30.
Page No 30.40:
Question 12:
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How many packets of each food should be used to minimise the amount of vitamin A in the diet? What is the minimum of vitamin A.
Answer:
Let x packets of food P and y packets of food Q be used to make the diet.
Each packet of food P contains 6 units of vitamin A and each packet of food Q contains 3 units of vitamin A. Therefore, x packets of food P and y packets of food Q contains (6x + 3y) units of vitamin A.
Since each packet of food P contains 12 units of calcium and each packet of food Q contains 3 units of calcium, therefore, x packets food P and y packets of food Q will contain (12x + 4y) units of calcium. But, the diet should contain atleast 240 units of calcium.
∴ 12x + 3y ≥ 240
⇒ 4x + y ≥ 80
Similarly, x packets of food P and y packets of food Q will contain (4x + 20y) units of iron. But, the diet should contain atleast 460 units of iron.
∴ 4x + 20y ≥ 460
⇒ x + 5y ≥ 115
Also, x packets of food P and y packets of food Q will contain (6x + 4y) units of cholesterol. But, the diet should contain atmost 300 units of cholesterol.
∴ 6x + 4y ≤ 300
⇒ 3x + 2y ≤ 150
Thus, the given linear programming problem is
Minimise Z = 6x + 3y
subject to the constraints
4x + y ≥ 80
x + 5y ≥ 115
3x + 2y ≤ 150
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(2, 72), B(15, 20) and C(40, 15).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 6x + 3y 
(2, 72)  6 × 0 + 3 × 72 = 216 
(15, 20)  6 × 15 + 3 × 20 = 150 
(40, 15)  6 × 40 + 3 × 15 = 285 
The smallest value of Z is 150 which is obtained at x = 15 and y = 20.
It can be verified that the open halfplane represented by 6x + 3y < 150 has no common points with the feasible region.
Thus, 15 packets of food P and 20 packets of food Q should be used to minimise the amount of vitamin A in the diet.
Hence, the minimum amount of vitamin A is 150 units.
Page No 30.40:
Question 13:
A farmer mixes two brands P and Q of cattle feed. Brand P, costing ₹250 per bag, contains 2 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Answer:
Let x bags of brand P and y bags of brand Q should be mixed to produce the mixture.
Each bag of brand P costs ₹250 and each bag of brand Q costs ₹200. Therefore, x bags of brand P and y bags of brand Q costs ₹(250x + 200y).
Since each bag of brand P contains 3 units of nutritional element A and each bag of brand Q contains 1.5 units of nutritional element A, therefore, x bag of brand P and y bag of brand Q will contain (3x + 1.5y) units of nutritional element A. But, the minimum requirement of nutrients A is 18 units.
∴ 3x + 1.5y ≥ 18
⇒ 2x + y ≥ 12
Similarly, x bag of brand P and y bag of brand Q will contain (2.5x + 11.25y) units of nutritional element B. But, the minimum requirement of nutrients B is 45 units.
∴ 2.5x + 11.25y ≥ 45
⇒ 2x + 9y ≥ 36
Also, x bag of brand P and y bag of brand Q will contain (2x + 3y) units of nutritional element B. But, the minimum requirement of nutrients C is 24 units.
∴ 2x + 3y ≥ 24
Thus, the given linear programming problem is
Minimise Z = 250x + 200y
subject to the constraints
2x + y ≥ 12
2x + 9y ≥ 36
2x + 3y ≥ 24
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(18, 0), B(9, 2), C(3, 6) and D(0, 12).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 250x + 200y 
(18, 0)  250 × 18 + 200 × 0 = 4500 
(9, 2)  250 × 9 + 200 × 2 = 2650 
(3, 6)  250 × 3 + 200 × 6 = 1950 → Minimum 
(0, 12)  250 × 0 + 200 × 12 = 2400 
The smallest value of Z is 1950 which is obtained at (3, 6).
It can be seen that the open halfplane represented by 250x + 200y < 1950 or 5x + 4y < 39 has no common points with the feasible region.
So, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost.
Hence, the minimum cost of the mixture per bag is ₹1950.
Page No 30.40:
Question 14:
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food  Vitamin A  Vitamin B  Vitamin C 
X  1  2  3 
Y  2  2  1 
One kg of food X costs ₹16 and one kg of food Y costs ₹20. Find the least cost of the mixture which will produce the required diet?
Answer:
Let x kg of food X and y kg of food Y are mixed together to make the mixture.
The cost of food X is ₹16/kg and that of food Y is ₹20/kg. So, x kg of food X and y kg of food Y will cost ₹(16x + 20y).
Since one kg of food X contains 1 unit of vitamin A and one kg of food Y contains 2 units of vitamin A, therefore, x kg of food X and y kg of food Y will contain (x + 2y) units of vitamin A. But, the mixture should contain atleast 10 units of vitamin A.
∴ x + 2y ≥ 10
Similarly, x kg of food X and y kg of food Y will contain (2x + 2y) units of vitamin B. But, the mixture should contain atleast 12 units of vitamin B.
∴ 2x + 2y ≥ 12
⇒ x + y ≥ 6
Also, x kg of food X and y kg of food Y will contain (3x + y) units of vitamin C. But, the mixture should contain atleast 8 units of vitamin C.
∴ 3x + y ≥ 8
Thus, the given linear programming problem is
Minimise Z = 16x + 20y
subject to the constraints
x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(10, 0), H(2, 4), G(1, 5) and F(0, 8).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 16x + 20y 
(10, 0)  16 × 10 + 20 × 0 = 160 
(2, 4)  16 × 2 + 20 × 4 = 112 → Minimum 
(1, 5)  16 × 1 + 20 × 5 = 116 
(0, 8)  16 × 0 + 20 × 8 = 160 
The smallest value of Z is 112 which is obtained at (2, 4).
It can be seen that the open halfplane represented by 16x + 20y < 112 has no common points with the feasible region.
So, the minimum value of Z is 112. Hence, the least cost of the mixture which will produce the required diet is ₹112.
Page No 30.40:
Question 15:
A fruit grower can use two types of fertilizer in his garden, brand P and Q. The amounts (in kg) of nirogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicates that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
kg per bag  
Brand P  Brand P  
Nitrogen  3  3.5 
Phosphoric acid  1  2 
Potash  3  1.5 
Chlorine  1.5  2 
If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
Answer:
Let x bags of fertilizer of brand P and y bags of fertilizer of brand Q are used in the garden.
One bag of brand P contains 3 kg of nitrogen and one bag of brand Q contains 3.5 kg of nitrogen, therefore, x bags of brand P and y bags of brand Q will contain (3x + 3.5y) kg of nitrogen.
Since one bag of brand P contains 1 kg of phosphoric acid and one bag of brand Q contains 2 kg of phosphoric acid, therefore, x bags of brand P and y bags of brand Q will contain (x + 2y) kg of phosphoric acid. But, the garden needs at least 240 kg of phosphoric acid.
∴ x + 2y ≥ 240
Similarly, x bags of brand P and y bags of brand Q will contain (3x + 1.5y) kg of potash. But, the garden needs at least 270 kg of potash.
∴ 3x + 1.5y ≥ 270
⇒ 2x + y ≥ 180
Also, x bags of brand P and y bags of brand Q will contain (1.5x + 2y) kg of chlorine. But, the garden needs at most 310 kg of chlorine.
∴ 1.5x + 2y ≤ 310
Thus, the given linear programming problem is
Minimise Z = 3x + 3.5y
subject to the constraints
x + 2y ≥ 240
2x + y ≥ 180
1.5x + 2y ≤ 310
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(40, 100), B(140, 50) and C(20, 140).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 3x + 3.5y 
(40, 100)  3 × 40 + 3.5 × 100 = 470 → Minimum 
(140, 50)  3 × 140 + 3.5 × 50 = 595 
(20, 140)  3 × 20 + 3.5 × 140 = 550 
The smallest value of Z is 470 which is obtained at (40, 100).
It can be seen that the open halfplane represented by 3x + 3.5y < 470 has no common points with the feasible region.
So, the minimum value of Z is 470.
Hence, 40 bags of fertilizer of brand P and 100 bags of fertilizer of brand Q are used in the garden. The minimum amount of nitrogen added in the garden is 470 kg.
Page No 30.50:
Question 1:
If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5/per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
Answer:
Let young man drives x km at a speed of 25 km/hr and y km at a speed of $40\mathrm{km}/\mathrm{hr}$.
Clearly, $x,y\ge 0$
It is given that, he spends Rs 2 per km if he drives at a speed of $25\mathrm{km}/\mathrm{hr}$ and Rs 5 per km if he drives at a speed of $40k\mathrm{m}/\mathrm{hr}$. Therefore, money spent by him when he travelled x km and y km is Rs 2x and Rs 5y respectively.
It is given that he has a maximum of Rs 100 to spend.
Thus, $2x+5y\le 100$
$\mathrm{Time}\mathrm{spent}\mathrm{by}\mathrm{him}\mathrm{when}\mathrm{travelling}\mathrm{with}\mathrm{a}\mathrm{speed}\mathrm{of}25\mathrm{km}/\mathrm{hr}=\frac{x}{25}\mathrm{hr}\phantom{\rule{0ex}{0ex}}\mathrm{Time}\mathrm{spent}\mathrm{by}\mathrm{him}\mathrm{when}\mathrm{travelling}\mathrm{with}\mathrm{a}\mathrm{speed}\mathrm{of}40\mathrm{km}/\mathrm{hr}=\frac{x}{40}\mathrm{hr}$
Also, the available time is of 1 hour.
$\frac{x}{25}+\frac{y}{40}\le 1\phantom{\rule{0ex}{0ex}}\Rightarrow 40x+25y\le 1000\phantom{\rule{0ex}{0ex}}$
The distance covered is Z = $x+y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $x+y$
subject to
$2x+5y\le 100\phantom{\rule{0ex}{0ex}}40x+25y\le 1000$
$x,y\ge 0$
First we will convert inequations into equations as follows:
2x + 5y = 100, 40x + 25y = 1000, x = 0 and y = 0
Region represented by 2x + 5y ≤ 100:
The line 2x + 5y = 100 meets the coordinate axes at $A\left(50,0\right)$ and $B\left(0,20\right)$ respectively. By joining these points we obtain the line
2x + 5y = 100. Clearly (0,0) satisfies the 2x + 5y = 100. So,the region which contains the origin represents the solution set of the inequation 2x + 5y ≤ 100.
Region represented by 40x + 25y ≤ 1000:
The line 40x + 25y = 1000 meets the coordinate axes at $C\left(25,0\right)$ and $D\left(0,40\right)$ respectively. By joining these points we obtain the line
2x + y = 12.Clearly (0,0) satisfies the inequation 40x + 25y ≤ 1000. So,the region which contains the origin represents the solution set of the inequation 40x + 25y ≤ 1000.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 5y ≤ 100, 40x + 25y ≤ 1000, x ≥ 0, and y ≥ 0 are as follows
The corner points are O(0, 0), B(0, 20), $E\left(\frac{50}{3},\frac{40}{3}\right)$ and C(25, 0).
The values of Z at these corner points are as follows
Corner point  Z = x + y 
O  0 
B  20 
E  30 
C  25 
The maximum value of Z is 30 which is attained at E.
Thus, the maximum distance travelled by the young man is 30 kms, if he drives $\frac{50}{3}\mathrm{km}$ at a speed of $25\mathrm{km}/\mathrm{hr}$ and $\frac{40}{3}\mathrm{km}$ at a speed of $40\mathrm{km}/\mathrm{hr}$ .
Page No 30.50:
Question 2:
A manufacturer has three machines installed in his factory. machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:
Item  Number of hours required by the machine  
A B 
I  II  III 
1 2 
2 1 
1 5/4 
Answer:
Let x units of item A and y units of item B be manufactured.
Therefore, $x,y\ge 0$
As we are given,
Item  Number of hours required by the machine  
A B 
I  II  III 
1 2 
2 1 
1 5/4 
Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day.
According to question, the constraints are
$x+2y\le 12\phantom{\rule{0ex}{0ex}}2x+y\le 12\phantom{\rule{0ex}{0ex}}x+\frac{5}{4}y\ge 5$
He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B.
Profit made by him in producing x items of A and y items of B is 6x + 4y.
Total profit Z = $6x+4y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $6x+4y$
subject to
$x+2y\le 12\phantom{\rule{0ex}{0ex}}2x+y\le 12\phantom{\rule{0ex}{0ex}}x+\frac{5}{4}y\ge 5$
$x,y\ge 0$
First we will convert inequations into equations as follows :
x + 2y = 12, 2x + y = 12, $x+\frac{5}{4}y=5$, x = 0 and y = 0
Region represented by x + 2y ≤ 12:
The line x + 2y = 12 meets the coordinate axes at A_{1}(12, 0) and B_{1}(0, 6) respectively. By joining these points we obtain the line x + 2y = 12.Clearly (0,0) satisfies the x + 2y = 12. So, the region which contains the origin represents the solution set of the inequation x + 2y ≤ 12.
Region represented by 2x + y ≤ 12:
The line 2x + y = 12 meets the coordinate axes at C_{1}(6, 0) and D_{1}(0, 12) respectively. By joining these points we obtain the line 2x + y = 12. Clearly (0,0) satisfies the inequation 2x + y ≤ 12. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 12.
Region represented by $x+\frac{5}{4}y\ge 5$:
The line $x+\frac{5}{4}y=5$ meets the coordinate axes at E_{1}(5, 0) and F_{1}(0, 4) respectively. By joining these points we obtain the line
$x+\frac{5}{4}y=5$. Clearly (0,0) does not satisfies the inequation $x+\frac{5}{4}y\ge 5$. So,the region which does not contains the origin represents the solution set of the inequation $x+\frac{5}{4}y\ge 5$.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + 2y ≤ 12, 2x + y ≤ 12, $x+\frac{5}{4}y\ge 5$, x ≥ 0, and y ≥ 0 are as follows.
The corner points are B_{1}(0, 6), G_{1}(4, 4), C_{1}(6, 0), E_{1}(5, 0) and F_{1}(0, 4).
The values of Z at these corner points are as follows
Corner point  Z = 6x + 4y 
B_{1}  24 
G_{1}  40 
C_{1}  36 
E_{1}  30 
F_{1}  16 
The maximum value of Z is 40 which is attained at G_{1}(4, 4).
Thus, the maximum profit is Rs 40 obtained when 4 units each of item A and B are manufactured.
Page No 30.50:
Question 3:
Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at a minimum labour cost?
Answer:
Let tailor A work for x days and tailor B work for y days.
In one day, A can stitch 6 shirts and 4 pants whereas B can stitch 10 shirts and 4 pants
Thus, in x days A can stitch $6x$ shirts and 4y pants whereas B can stich
pants.
It is given that the minimum requirement of the shirts and pants are respectively 60 and 32.
Thus,
$6x+10y\ge 60\phantom{\rule{0ex}{0ex}}4x+4y\ge 32$
Further it is given that A and B earn Rs 15 and Rs 20 per day respectively.
Thus, A earn Rs 15x and B earn Rs 20y .
Let Z denotes the total cost
$\therefore Z=15x+20y$
Days cannot be negative.
$\therefore $$x,y\ge 0$
Min Z = $15x+20y$
subject to
$6x+10y\ge 60\phantom{\rule{0ex}{0ex}}4x+4y\ge 32$
$x,y\ge 0$
First we will convert inequations into equations as follows :
6x + 10y = 60, 4x + 4y = 32, x = 0 and y = 0
Region represented by 6x + 10y ≥ 60:
The line 6x + 10y = 60 meets the coordinate axes at A_{1}(10, 0) and B_{1}(0, 6) respectively. By joining these points we obtain the line6x + 10y = 60. Clearly (0,0) does not satisfies the 6x + 10y = 60. So,the region which does not contains the origin represents the solution set of the inequation 6x + 10y ≥ 60.
Region represented by 4x + 4y ≥ 32:
The line 4x + 4y =32 meets the coordinate axes at C_{1}(8, 0) and D_{1}(0, 8) respectively. By joining these points we obtain the line 4x + 4y = 32.Clearly (0,0) does not satisfies the inequation 4x + 4y ≥ 32. So,the region which does not contains the origin represents the solution set of the inequation 4x + 4y ≥ 32.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 6x + 10y ≥ 60,4x + 4y ≥ 32, x ≥ 0, and y ≥ 0 are as follows.
Thus, the mathematical formulation of the given linear programming problem is
The corner points are D_{1}(0, 8), E_{1}(5, 3) and A_{1}(10, 0).
The values of Z at these corner points are as follows
Corner point  Z = 15x + 20y 
D_{1}  160 
E_{1}  135 
A_{1}  150 
The minimum value of Z is 135 which is attained at E_{1}(5, 3).
Thus, for minimum labour cost, A should work for 5 days and B should work for 3 days.
Page No 30.50:
Question 4:
A factory manufactures two types of screws, A and B, each type requiring the use of two machines  an automatic and a handoperated. It takes 4 minute on the automatic and 6 minutes on the handoperated machines to manufacture a package of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the handoperated machine to manufacture a package of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws 'A' at a profit of 70 P and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Answer:
Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.
Screw A  Screw B  Availability  
Automatic Machine (min)  4  6  4 × 60 = 240 
Hand Operated Machine (min)  6  3  4 × 60 = 240 
The manufacturer can sell a package of screws 'A' at a profit of Rs 0.7 and screws 'B' at a profit of Re 1.
Total profit, Z = 0.7x + 1y
The mathematical formulation of the given problem is
Maximize Z = 0.7x + 1y
subject to the constraints,
x, y ≥ 0
First we will convert inequations into equations as follows:
4x + 6y = 240, 6x + 3y = 240, x = 0 and y = 0
Region represented by 4x + 6y ≤ 240:
The line 4x + 6y = 240 meets the coordinate axes at A_{1}(60, 0) and B_{1}(0, 40) respectively. By joining these points we obtain the line 4x + 6y = 240.Clearly (0,0) satisfies the 4x + 6y = 240. So,the region which contains the origin represents the solution set of the inequation 4x + 6y ≤ 240.
Region represented by 6x + 3y ≤ 240:
The line 6x + 3y = 240 meets the coordinate axes at C_{1}(40, 0) and D_{1}(0, 80) respectively. By joining these points we obtain the line 6x + 3y = 240. Clearly (0,0) satisfies the inequation 6x + 3y ≤ 240. So,the region which contains the origin represents the solution set of the inequation 6x + 3y ≤ 240.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 4x + 6y ≤ 240, 6x + 3y ≤ 240, x ≥ 0, and y ≥ 0 are as follows.
The corner points are C_{1}(40, 0), E_{1}(30, 20) and B_{1}(0, 40).
The values of Z at these corner points are as follows.
Corner point  Z = 7x + 10y 
C_{1}(40, 0)  280 
E_{1}(30, 20)  410 
B_{1}(0, 40)  400 
The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
Page No 30.50:
Question 5:
A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of a lower quality. Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day.
How should the company manufacture the two types of belts in order to have a maximum overall profit?
Answer:
Let the company produces x belts of type A and y belts of type B.
Number of belts cannot be negative.
Therefore, $x,y\ge 0$
It is given that leather is sufficient only for 800 belts per day (both A and B combined). Therefore,
x + y ≤ 800
It is given that the rate of production of belts of type B is 1000 per day.Hence, the time taken to produce y belts of type B is $\frac{y}{1000}$.
And, since each belt of type A requires twice as much time as a belt of type B, the rate of production of belts of type A is 500 per day and therefore, total time taken to produce x belts of type A is $\frac{x}{500}$.Thus, we have
$\frac{x}{500}+\frac{y}{1000}\le 1\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+y\le 1000$
Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day.
x ≤ 400
For belt of type B, only 700 buckles are available per day.
y ≤ 700
Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Therefore, profit gained on x belts of type A and y belts of type B is Rs 2x and
Rs 1.50y respectively.Hence, the total profit would be Rs (2x + 1.50y).
Let Z denote the total profit.
$\therefore $ Z = $2x+1.5y$
Thus, the mathematical formulation of the given linear programming problem is
Max Z = $2x+1.5y$
subject to
$x+y\le 800\phantom{\rule{0ex}{0ex}}2x+y\le 1000\phantom{\rule{0ex}{0ex}}x\le 400\phantom{\rule{0ex}{0ex}}y\le 700$
$x,y\ge 0$
First we will convert inequations into equations as follows :
x + y = 800, 2x + y = 1000, x = 400, y = 700, x = 0 and y = 0
Region represented by x + y ≤ 800:
The line x + y = 800 meets the coordinate axes at A_{1}(800, 0) and B_{1}(0, 800) respectively. By joining these points we obtain the line x + y = 800. Clearly (0,0) satisfies the x + y = 800. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 800.
Region represented by 2x + y ≤ 1000:
The line 2x + y = 1000 meets the coordinate axes at C_{1}(500, 0) and D_{1}(0, 1000) respectively. By joining these points we obtain the line 2x + y = 1000. Clearly (0,0) satisfies the inequation 2x + y ≤ 1000. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 1000.
Region represented by x ≤ 400:
The line x = 400 will pass through E_{1}(400, 0). The region to the left of the line x = 400 will satisfy the inequation x ≤ 400.
Region represented by y ≤ 700:
The line y = 700 will pass through F_{1}(0, 700). The region below the line y = 700 will satisfy the inequation y ≤ 700.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 800, 2x + y ≤ 1000, x ≤ 400, y ≤ 700, x ≥ 0, and y ≥ 0 are as follows.
The feasible region determined by the system of constraints is
The corner points are F_{1}(0, 700), G_{1}(200, 600), H_{1}(400, 200) and E_{1}(400, 0).
The values of Z at these corner points are as follows
Corner point  Z= 2x +1.5y 
F_{1}(0, 700)  1050 
G_{1}(200, 600)  1300 
H_{1}(400, 200)  1100 
E_{1}(400, 0)  800 
The maximum value of Z is 1300 which is attained at G_{1}(200, 600).
Thus, the maximum profit is Rs 1300 obtained when 200 belts of type A and 600 belts of type B were produced.
Page No 30.51:
Question 6:
A small manufacturer has employed 5 skilled men and 10 semiskilled men and makes an article in two qualities deluxe model and an ordinary model. The making of a deluxe model requires 2 hrs. work by a skilled man and 2 hrs. work by a semiskilled man. The ordinary model requires 1 hr by a skilled man and 3 hrs. by a semiskilled man. By union rules no man may work more than 8 hrs per day. The manufacturers clear profit on deluxe model is Rs 15 and on an ordinary model is Rs 10. How many of each type should be made in order to maximize his total daily profit.
Answer:
Let x articles of deluxe model and y articles of an ordinary model be made.
Number of articles cannot be negative.
Therefore, $x,y\ge 0$
According to the question, the making of a deluxe model requires 2 hrs. work by a skilled man and the ordinary model requires 1 hr by a skilled man
$2x+y\le 40\phantom{\rule{0ex}{0ex}}$
The making of a deluxe model requires 2 hrs. work by a semiskilled man ordinary model requires 3 hrs. work by a semiskilled man.
$2x+3y\le 80$
Total profit = Z = $15x+10y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $15x+10y$
subject to
$2x+y\le 40\phantom{\rule{0ex}{0ex}}2x+3y\le 80$
$x\ge 0\phantom{\rule{0ex}{0ex}}y\ge 0$
The feasible region determined by the system of constraints is
The corner points are A(0,$\frac{80}{3}$), B(10, 20), C(20, 0)
The values of Z at these corner points are as follows
Corner point  Z= 15x+10y 
A  $\frac{800}{3}$ 
B  350 
C  300 
The maximum value of Z is 300 which is attained at C(20, 0)
Thus, the maximum profit is Rs 300 obtained when 10 units of deluxe model and 20 unit of ordinary model is produced
Page No 30.51:
Question 7:
A manufacturer makes two types A and B of teacups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:
Machines  
I  II  III  
A B 
12 6 
18 0 
6 9 
Answer:
Let x units of type A and y units of type B cups were made.
Quantities cannot be negative.Therefore,
$x,y\ge 0$
As we are given,
Machines  
I  II  III  
A B 
12 6 
18 0 
6 9 
Every machine is available for a maximum of 6 hours per day i.e. 360 minutes per day.
Therefore, the constraints are
$12x+6y\le 360\phantom{\rule{0ex}{0ex}}18x+0y\le 360\phantom{\rule{0ex}{0ex}}6x+9y\le 360\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
If the profit on each cup A is 75 paise and that on each cup B is 50 paise.
Total profit = Z = $0.75x+0.50y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $0.75x+0.50y$
subject to
$12x+6y\le 360\phantom{\rule{0ex}{0ex}}18x+0y\le 360\phantom{\rule{0ex}{0ex}}6x+9y\le 360\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$x,y\ge 0$
First we will convert inequations into equations as follows :
12x + 6y = 360, 18x = 360, $6x+9y=360$, x = 0 and y = 0
Region represented by 12x + 6y ≤ 360:
The line 12x + 6y = 360 meets the coordinate axes at A_{1}(30, 0) and B_{1}(0, 60) respectively. By joining these points we obtain the line 12x + 6y = 360.Clearly (0,0) satisfies the 12x + 6y = 360. So,the region which contains the origin represents the solution set of the inequation 12x + 6y ≤ 360.
Region represented by 18x + 0y ≤ 360:
The line 18x + 0y = 360 meets the coordinate axes at C_{1}(20, 0) . We obtain the line 18x + 0y = 360.Clearly (0,0) satisfies the inequation 18x + 0y ≤ 360. So,the region which contains the origin represents the solution set of the inequation 18x + 0y ≤ 360.
Region represented by $6x+9y\le 360$:
The line $6x+9y=360$ meets the coordinate axes at E_{1}(60, 0) and F_{1}(0, 40) respectively. By joining these points we obtain the line $6x+9y=360$. Clearly (0,0) satisfies the inequation $6x+9y\le 360$ . So,the region which contains the origin represents the solution set of the inequation $6x+9y\le 360$.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 12x + 6y ≤ 360, 18x + 0y ≤ 360, $6x+9y\le 360$, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0) F_{1}(0, 40), G_{1}(15, 30), H_{1}(20, 20) and C_{1}(20, 0).
The values of Z at these corner points are as follows
Corner point  Z= 0.75x + 0.50y 
O  0 
F_{1}  20 
G_{1}  26.25 
H_{1}  25 
C_{1}  15 
Thus, the maximum profit is Rs 26.25 obtained when 15 units of type A and 30 units of type B cups were made.
Page No 30.51:
Question 8:
A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:
Area occupied by the machine 
Labour force for each machine 
Daily output in units 

Machine A Machine B 
1000 sq. m 1200 sq. m 
12 men 8 men 
60 40 
How many machines of each type should he buy to maximize the daily output?
Answer:
Let x machines of type A and y machines of type B were purchased.
Number of machines cannot be negative.
Therefore, $x,y\ge 0$
We are given,
Area occupied by the machine 
Labour force for each machine 
Daily output in units 

Machine A Machine B 
1000 sq. m 1200 sq. m 
12 men 8 men 
60 40 
The area of 7600 sq m is available and there are 72 skilled men available to operate the machines.
Therefore, the constraints are
$1000x+1200y\le 7600\phantom{\rule{0ex}{0ex}}\mathrm{and}12x+8y\le 72$
Total daily output = Z = $60x+40y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programming problem is
Max Z = $60x+40y$
subject to
$1000x+1200y\le 7600\phantom{\rule{0ex}{0ex}}12x+8y\le 72$
$x,y\ge 0$
First we will convert inequations into equations as follows :
1000x + 1200y = 7600, 12x + 8y = 72, x = 0 and y = 0
Region represented by 1000x + 1200y ≤ 7600:
The line 1000x + 1200y = 7600 meets the coordinate axes at ${A}_{1}\left(\frac{38}{5},0\right)$ and ${B}_{1}\left(0,\frac{19}{3}\right)$ respectively. By joining these points we obtain the line
1000x + 1200y = 7600. Clearly (0,0) satisfies the 1000x + 1200y = 7600. So, the region which contains the origin represents the solution set of the inequation 1000x + 1200y ≤ 7600.
Region represented by 12x + 8y ≤ 72:
The line 12x + 8y = 72 meets the coordinate axes at C_{1}(6, 0) and D_{1}(0, 9) respectively. By joining these points we obtain the line 12x + 8y = 72 .Clearly (0,0) satisfies the inequation 12x + 8y ≤ 72. So,the region which contains the origin represents the solution set of the inequation 12x + 8y ≤ 72.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 1000x + 1200y ≤ 7600, 12x + 8y ≤ 72, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0) ${B}_{1}\left(0,\frac{19}{3}\right)$, E_{1}(4, 3), C_{1}(6, 0)
The values of Z at these corner points are as follows
Corner point  Z= 60x + 40y 
O  0 
B_{1}  253.3 
E_{1}  360 
C_{1}  360 
The maximum value of Z is 360 which is attained at E_{1}(4, 3) and C_{1}(6, 0).
Thus, the maximum output is Rs 360 obtained when 4 units of type A and 3 units of type B or 6 units of type A are manufactured.
Page No 30.51:
Question 9:
A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 gm of silver and 1 gm of gold while that of type B requires 1 gm of silver and 2 gm of gold. The company can produce 9 gm of silver and 8 gm of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?
Answer:
Let x goods of type A and y goods of type B were produced.
Number of goods cannot be negative.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Silver(gm)  Gold white(gm)  
Type A  3  1 
Type B  1  2 
Availability  9  8 
Therefore, the constraints are
$3x+y\le 9\phantom{\rule{0ex}{0ex}}x+2y\le 8$
If each unit of type A brings a profit of Rs 40 and that of type B Rs 50.Then, x goods of type A and y goods of type B brings a profit of Rs 40x and Rs 50y.
Total profit = Z = $40x+50y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $40x+50y$
subject to
$3x+y\le 9\phantom{\rule{0ex}{0ex}}x+2y\le 8$
$x,y\ge 0$
First we will convert inequations into equations as follows :
3x + y = 9, x + 2y = 8, x = 0 and y = 0
Region represented by 3x + y ≤ 9:
The line 3x + y = 9 meets the coordinate axes at A_{1}(3, 0) and B_{1}(0, 9) respectively. By joining these points we obtain the line
3x + y = 9.Clearly (0,0) satisfies the 3x + y = 9. So,the region which contains the origin represents the solution set of the inequation
3x + y ≤ 9.
Region represented by x + 2y ≤ 8:
The line x + 2y = 8 meets the coordinate axes at C_{1}(8, 0) and D_{1}(0, 4) respectively. By joining these points we obtain the line
x + 2y = 8.Clearly (0,0) satisfies the inequation x + 2y ≤ 8. So,the region which contains the origin represents the solution set of the inequation x + 2y ≤ 8.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0), D_{1}(0, 4), E_{1}(2, 3), A_{1}(3, 0)
The values of Z at these corner points are as follows
Corner point  Z = 40x + 50y 
O  0 
D_{1}  200 
E_{1}  230 
A_{1}  120 
The maximum value of Z is 230 which is attained at E_{1}(2, 3).
Thus, the maximum profit is of Rs 230 obtained when 2 units of type A and 3 units of type B produced.
Page No 30.51:
Question 10:
A manufacturer of Furniture makes two products : chairs and tables. processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his profit.
Answer:
Let x chairs and y tables were produced.
Number of chairs and tables cannot be negative.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Time on machine A(hrs)  Time on machine B (hrs)  
Chairs  2  6 
Tables  4  2 
Availability  16  30 
Therefore, the constraints are
$2x+4y\le 16\phantom{\rule{0ex}{0ex}}6x+2y\le 30\phantom{\rule{0ex}{0ex}}$
Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively.Therefore, profit gained from x chairs and y tables is Rs 3x and Rs 5y.
Total profit = Z = $3x+5y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $3x+5y$
subject to
$2x+4y\le 16\phantom{\rule{0ex}{0ex}}6x+2y\le 30\phantom{\rule{0ex}{0ex}}$
$x,y\ge 0$
First we will convert inequations into equations as follows:
2x + 4y = 16, 6x + 2y =30, x = 0 and y = 0
Region represented by 2x + 4y ≤ 16:
The line 2x + 4y = 16 meets the coordinate axes at A_{1}(8, 0) and B_{1}(0, 4) respectively. By joining these points we obtain the line 2x + 4y = 16. Clearly (0,0) satisfies the 2x + 4y = 16. So, the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 16.
Region represented by 6x + 2y ≤ 30:
The line 6x + 2y =30 meets the coordinate axes at C_{1}(5, 0) and D_{1}(0, 15) respectively. By joining these points we obtain the line 6x + 2y =30 .Clearly (0,0) satisfies the inequation 6x + 2y ≤ 30. So,the region which contains the origin represents the solution set of the inequation 6x + 2y ≤ 30.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 4y ≤ 16, 6x + 2y ≤ 30, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0), B_{1}(0, 4), E_{1}$\left(\frac{22}{5},\frac{9}{5}\right)$ and C_{1}(5, 0).
The values of Z at these corner points are as follows
Corner point  Z= 3x + 5y 
O  0 
B_{1}  20 
E_{1}  22.2 
C_{1}  15 
The maximum value of Z is 22.2 which is attained at B_{1}$\left(\frac{22}{5},\frac{9}{5}\right)$.
Thus, the maximum profit is of Rs 22.20 obtained when $\frac{22}{5}$ units of chairs and $\frac{9}{5}$ units of tables are produced.
Page No 30.51:
Question 11:
A furniture manufacturing company plans to make two products : chairs and tables. From its available resources which consists of 400 square feet to teak wood and 450 man hours. It is known that to make a chair requires 5 square feet of wood and 10 manhours and yields a profit of Rs 45, while each table uses 20 square feet of wood and 25 manhours and yields a profit of Rs 80. How many items of each product should be produced by the company so that the profit is maximum?
Answer:
Let x units of chairs and y units of tables were produced
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Wood(square feet)  Man hours  
Chairs(x)  5  10 
Tables(y)  20  25 
Availability  400  450 
Therefore, the constraints are
$5x+20y\le 400\phantom{\rule{0ex}{0ex}}10x+25y\le 450\phantom{\rule{0ex}{0ex}}$
It is known that to make a chair requires 5 square feet of wood and 10 manhours and yields a profit of Rs 45, while each table uses 20 square feet of wood and 25 manhours and yields a profit of Rs 80.
Therefore, profit gained to make x chairs and y tables is Rs 45x and Rs 80y respectively.
Total profit = Z = $45x+80y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $45x+80y$
subject to
$5x+20y\le 400\phantom{\rule{0ex}{0ex}}10x+25y\le 450\phantom{\rule{0ex}{0ex}}$
$x,y\ge 0$
First we will convert inequations into equations as follows:
5x + 20y = 400, 10x + 25y = 450, x = 0 and y = 0
Region represented by 5x + 20y ≤ 400:
The line 5x + 20y = 400 meets the coordinate axes at $A\left(80,0\right)$ and $B\left(0,20\right)$ respectively. By joining these points we obtain the line
5x + 20y = 400 . Clearly (0,0) satisfies the 5x + 20y = 400 . So, the region which contains the origin represents the solution set of the inequation 5x + 20y ≤ 400.
Region represented by 10x + 25y ≤ 450:
The line 10x + 25y = 450 meets the coordinate axes at $C\left(45,0\right)$ and $D\left(0,18\right)$ respectively. By joining these points we obtain the line
10x + 25y = 450. Clearly (0,0) satisfies the inequation 10x + 25y ≤ 450. So,the region which contains the origin represents the solution set of the inequation 10x + 25y ≤ 450.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + 20y ≤ 400, 10x + 25y ≤ 450, x ≥ 0, and y ≥ 0 are as follows.
The corner points are A(0, 18), B(45, 0)
The values of Z at these corner points are as follows
Corner point  Z= 45x + 80y 
A  1440 
B  2025 
The maximum value of Z is 2025 which is attained at B$\left(45,0\right)$.
Thus, the maximum profit is of Rs 2025 obtained when 45 units of chairs and no units of tables are produced
Page No 30.52:
Question 12:
A firm manufactures two products A and B. Each product is processed on two machines M_{1} and M_{2}. Product A requires 4 minutes of processing time on M_{1} and 8 min. on M_{2} ; product B requires 4 minutes on M_{1} and 4 min. on M_{2}. The machine M_{1} is available for not more than 8 hrs 20 min. while machine M_{2} is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively.
Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.
Answer:
Let x products of type A and y products of type B are manufactured.
Number of products cannot be negative.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Product  ${M}_{1}$(minutes)  ${M}_{2}$(minutes) 
A(x)  4  8 
B(y)  4  4 
Availability  500  600 
Therefore, the constraints are
$4x+4y\le 500\phantom{\rule{0ex}{0ex}}8x+4y\le 600$
The products A and B are sold at a profit of Rs 3 and Rs 4 respectively. Therefore, Profit gained from x products of type A and y products of type B is Rs 3x and Rs 4y respectively.
Total profit = Z = 3x + 4y which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $3x+4y$
subject to
$4x+4y\le 500\phantom{\rule{0ex}{0ex}}8x+4y\le 600$
First we will convert inequations into equations as follows:
4x + 4y = 500, 8x + 4y = 600, x = 0 and y = 0
Region represented by 4x + 4y ≤ 500:
The line 4x + 4y = 500 meets the coordinate axes at A_{1}(125, 0) and B_{1}(0, 125) respectively. By joining these points we obtain the line
4x + 4y = 500. Clearly (0,0) satisfies the 5x + 20y = 400 . So, the region which contains the origin represents the solution set of the inequation 5x + 20y ≤ 400.
Region represented by 8x + 4y ≤ 600:
The line 8x + 4y = 600 meets the coordinate axes at C_{1}(75, 0) and D_{1}(0, 150) respectively. By joining these points we obtain the line
8x + 4y = 600. Clearly (0,0) satisfies the inequation 8x + 4y ≤ 600. So,the region which contains the origin represents the solution set of the inequation
8x + 4y ≤ 600.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 4x + 4y ≤ 500, 8x + 4y ≤ 600, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0), B_{1}(0, 125), E_{1}(25, 100) and C_{1}(75, 0).
The values of Z at these corner points are as follows
Corner point  Z= 3x + 4y 
O  0 
B_{1}  500 
E_{1}  475 
C_{1}  225 
The maximum value of Z is 500 which is attained at B_{1}(0, 125).
Thus, the maximum profit is Rs 500 obtained when no units of product A and 125 units of product B were manufactured.
Page No 30.52:
Question 13:
A firm manufacturing two types of electric items, A and B, can make a profit of Rs 20 per unit of A and Rs 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.
Answer:
Let x units of item A and y units of item B were manufactured.
Number of items cannot be negative.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Product  Motors  Transformers 
A(x)  3  4 
B(y)  2  4 
Availability  210  300 
Further, it is given that type B is an export model, whose supply is restricted to 65 units per month.
Therefore, the constraints are
$3x+2y\le 210\phantom{\rule{0ex}{0ex}}4x+4y\le 300\phantom{\rule{0ex}{0ex}}y\le 65$
A and B can make a profit of Rs 20 per unit of A and Rs 30 per unit of B.Therefore, profit gained from x units of item A and y units of item B is Rs 20x and Rs 30y respectively.
Total profit = Z = $20x+30y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $20x+30y$
subject to
$3x+2y\le 210\phantom{\rule{0ex}{0ex}}4x+4y\le 300\phantom{\rule{0ex}{0ex}}y\le 65$
$x,y\ge 0$
First we will convert inequations into equations as follows:
3x + 2y = 210, 4x + 4y = 300, y = 65, x = 0 and y = 0
Region represented by 3x + 2y ≤ 210:
The line 3x + 2y = 210 meets the coordinate axes at A_{1}(70, 0) and B_{1}(0, 105) respectively. By joining these points we obtain the line 3x + 2y = 210.Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 3x + 2y ≤ 210.
Region represented by 4x + 4y ≤ 300:
The line 4x + 4y = 300 meets the coordinate axes at C_{1}(75, 0) and D_{1}(0, 75) respectively. By joining these points we obtain the line
4x + 4y = 300.Clearly (0,0) satisfies the inequation 4x + 4y ≤ 300. So,the region which contains the origin represents the solution set of the inequation
4x + 4y ≤ 300.
y = 65 is the line passing through the point E_{1}(0, 65) and is parallel to X axis.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 3x + 2y ≤ 210, 4x + 4y ≤ 300, y ≤ 65, x ≥ 0, and y ≥ 0 are as follows
The corner points are O(0, 0), E_{1}(0, 65), G_{1}$\left(10,65\right)$, F_{1}(60, 15) and A_{1}(70, 0).
The values of Z at these corner points are as follows
Corner point  Z= 20x + 30y 
O  0 
E_{1}  1950 
G_{1}  2150 
F_{1}  1650 
A_{1}  1400 
The maximum value of Z is 2150 which is attained at G_{1}$\left(10,65\right)$.
Thus, the maximum profit is Rs 2150 obtained when 10 units of item A and 65 units of item B were manufactured.
Page No 30.52:
Question 14:
A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
Answer:
Let x units of first product and y units of second product be manufactured.
Therefore, $x,y\ge 0$
The given information can be tabulated as follows:
Product  Resource A  Resource B  Resource C 
First(x)  2  2  4 
Second(y)  4  2  0 
Availability  20  12  16 
Therefore, the constraints are
$2x+4y\le 20\phantom{\rule{0ex}{0ex}}2x+2y\le 12\phantom{\rule{0ex}{0ex}}4x+0y\le 16\mathrm{or}4x\le 16\phantom{\rule{0ex}{0ex}}$
It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Therefore, profit gained from x units of first product and y units of second product is 2x monetary units and 4y monetary units respectively.
Total profit = Z = $2x+3y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $2x+3y$
subject to
$2x+4y\le 20\phantom{\rule{0ex}{0ex}}2x+2y\le 12\phantom{\rule{0ex}{0ex}}4x+0y\le 16\mathrm{or}4x\le 16\phantom{\rule{0ex}{0ex}}$
$x,y\ge 0$
First we will convert inequations into equations as follows :
2x + 4y = 20, 2x + 2y = 12, 4x = 16, x = 0 and y = 0
Region represented by 2x + 4y ≤ 20:
The line 2x + 4y = 20 meets the coordinate axes at A_{1}(10, 0) and B_{1}(0, 5) respectively. By joining these points we obtain the line 2x + 4y = 20.Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 20.
Region represented by 2x + 2y ≤ 12:
The line 2x +2y =16 meets the coordinate axes at C_{1}(6, 0) and D_{1}(0, 6) respectively. By joining these points we obtain the line 2x + 2y = 12.Clearly (0,0) satisfies the inequation 2x + 2y ≤ 12. So, the region which contains the origin represents the solution set of the inequation 2x + 2y ≤ 12.
Region represented by 4x ≤ 16:
The line 4x =16 or x = 4 is the line passing through the point E_{1}(4, 0) and is parallel to Y axis.The region to the left of the line x = 4 would satisfy the inequation 4x ≤ 16.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 4y ≤ 20, 2x + 2y ≤ 12, 4x ≤ 16, x ≥ 0 and y ≥ 0 are as follows
The corner points are O(0, 0), B_{1}(0, 5), G_{1}$\left(2,4\right)$, F_{1}(4, 2) and E_{1}(4, 0).
The values of Z at these corner points are as follows
Corner point  Z= 2x + 3y 
O  0 
B_{1}  15 
G_{1}  16 
F_{1}  14 
E_{1}  8 
The maximum value of Z is 16 which is attained at G_{1}$\left(2,4\right)$.
Thus, the maximum profit is 16 monetary units obtained when 2 units of first product and 4 units of second product were manufactured.
Page No 30.52:
Question 15:
A publisher sells a hard cover edition of a text book for Rs 72.00 and paperback edition of the same ext for Rs 40.00. Costs to the publisher are Rs 56.00 and Rs 28.00 per book respectively in addition to weekly costs of Rs 9600.00. Both types require 5 minutes of printing time, although hardcover requires 10 minutes binding time and the paperback requires only 2 minutes. Both the printing and binding operations have 4,800 minutes available each week. How many of each type of book should be produced in order to maximize profit?
Answer:
Let the sale of hand cover edition be 'h' and that of paperback editions be 't'.
SP of a hard cover edition of the textbook = Rs 72
SP of a paperback edition of the textbook = Rs 40
Cost to the publisher for hard cover edition = Rs 56
Cost to the publisher for a paperback edition = Rs 28
Weekly cost to the publisher = Rs 9600
Profit to be maximised, Z = (72 − 56) h + (40 − 28) t − 9600
⇒ Z = 16h + 12t − 9600
5 (h + t) ≤ 4800
10 h + 2t ≤ 4800
The corner points are O(0, 0), B_{1}(0, 960), E_{1}(360, 600) and F_{1}(480, 0).
The values of Z at these corner points are as follows:
Corner point  Z = 16h + 12t − 9600 
O  −9600 
B_{1}  1920 
E_{1}  3360 
F_{1}  −1920 
The maximum value of Z is 3360 which is attained at E_{1}(360, 600).
The maximum profit is 3360 which is obtained by selling 360 copies of hardcover edition and 600 copies of paperback edition.
Page No 30.52:
Question 16:
A firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine; size B contains 1 grain of aspirin, 8 grains of bicarbonate and 66 grains of codeine. It has been found by users that it requires at least 12 grains of aspirin, 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effects. Determine graphically the least number of pills a patient should have to get immediate relief. Determine also the quantity of codeine consumed by patient.
Answer:
Let the number of size A pill be x and the number of size B pill be y.
Therefore, the constraints are
$\begin{array}{lc}2x+y\ge 12& \\ 5x+8y\ge 7.4& \\ x+66y\ge 24& \end{array}$
Z = $x+y$ which is to be minimised
The corner points are (0, 12), (24, 0) and $\left(\frac{768}{131},\frac{36}{131}\right)$.
The values of Z at these corner points are as follows
Corner point  Z = x + y 
(0, 12)  12 
(24, 0)  24 
$\left(\frac{768}{131},\frac{36}{131}\right)$  6.1373 
The minimum value of Z is 6.1373 but the region is unbounded so check whether x + y < 6.1373 has common region with the feasible solution.
Clearly, it can be seen that it doesn't has any common region.
So,
$x=\frac{768}{131},y=\frac{36}{131}$
This is the least quantity of pill A and B.
$\begin{array}{l}\mathrm{Codline}\mathrm{quantity}\Rightarrow \frac{768}{131}+66\times \frac{36}{131}\\ \Rightarrow 24\hspace{0.17em}\mathrm{grains}\end{array}$
Disclaimer: The answer provided in the book is incorrect.
Page No 30.52:
Question 17:
A chemical company produces two compounds, A and B. The following table gives the units of ingredients, C and D per kg of compounds A and B as well as minimum requirements of C and D and costs per kg of A and B. Find the quantities of A and B which would give a supply of C and D at a minimum cost.
Compound  Minimum requirement  
A  B  
Ingredient C Ingredient D 
1 3 
2 1 
80 75 
Cost (in Rs) per kg  4  6 
Answer:
Let x kg of compound A and y kg of compound B were produced.
Quantity cannot be negative.
Therefore, $x,y\ge 0$
Compound  Minimum requirement  
A  B  
Ingredient C Ingredient D 
1 3 
2 1 
80 75 
Cost (in Rs) per kg  4  6 
According to question, the constraints are
$x+2y\ge 80\phantom{\rule{0ex}{0ex}}3x+y\ge 75$
Cost (in Rs) per kg of compound A and compound B is Rs 4 and Rs 6 respectively.Therefore, cost of x kg of compound A and y kg of compound B is 4x and 6y respectively.
Total cost = Z = $4x+6y$ which is to be minimised.
Thus, the mathematical formulation of the given linear programmimg problem is
Min Z = $4x+6y$
subject to
$x+2y\ge 80\phantom{\rule{0ex}{0ex}}3x+y\ge 75$
$x,y\ge 0$
First we will convert inequations into equations as follows:
x + 2y = 80, 3x + y =75, x = 0 and y = 0
Region represented by x + 2y ≥ 80:
The line x + 2y = 80 meets the coordinate axes at A_{1}(80, 0) and B_{1}(0, 40) respectively. By joining these points we obtain the line x + 2y = 80. Clearly (0,0) does not satisfies the x + 2y = 80. So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 80.
Region represented by 3x + y ≥ 75:
The line 3x + y =75 meets the coordinate axes at C_{1}(25, 0) and D_{1}(0, 75) respectively. By joining these points we obtain the line 3x + y =75.Clearly (0,0) does not satisfies the inequation 3x + y ≥ 75. So,the region which does not contain the origin represents the solution set of the inequation 3x + y ≥ 75.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, and y ≥ 0 are as follows.
The corner points are D_{1}(0, 75), E_{1}(14, 33) and A_{1}(80, 0).
The values of Z at these corner points are as follows
Corner point  Z= 4x + 6y 
D_{1}  450 
E_{1}  254 
A_{1}  320 
The minimum value of Z is 254 which is attained at E_{1}$\left(14,33\right)$.
Thus, the minimum cost is Rs 254 obtained when 14 units of compound A and 33 units of compound B were produced.
Page No 30.52:
Question 18:
A company manufactures two types of novelty Souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?
Answer:
Let the company manufacture x souvenirs of type A and y souvenirs of type B.
Number of items cannot be negative.
Therefore,
x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.
Type A 
Type B 
Availability 

Cutting(min) 
5 
8 
3 × 60 + 20 = 200 
Assembling(min) 
10 
8 
4 × 60 = 240 
Therefore, the constraints are
The profit on type A souvenirs is 50 paise and on type B souvenirs is 60 paise.Therefore, profit gained on x souvenirs of type A and y souvenirs of type B is Rs 0.50x and Rs 0.60y respectively.
Total profit, Z = 0.5x + 0.6y
The mathematical formulation of the given problem is
Maximize Z = 0.5x + 0.6y
subject to the constraints,
x ≥ 0 and y ≥ 0
First we will convert inequations into equations as follows:
5x + 8y = 200, 10x + 8y = 240, x = 0 and y = 0
Region represented by 5x + 8y ≤ 200:
The line 5x + 8y = 200 meets the coordinate axes at A_{1}(40, 0) and B_{1}(0, 25) respectively. By joining these points we obtain the line
5x + 8y = 200 . Clearly (0,0) satisfies the 5x + 8y = 200. So, the region which contains the origin represents the solution set of the inequation 5x + 8y ≤ 200.
Region represented by 10x + 8y ≤ 240:
The line 10x + 8y = 240 meets the coordinate axes at C_{1}(24, 0) and D_{1}(0, 30) respectively. By joining these points we obtain the line
10x + 8y = 240. Clearly (0,0) satisfies the inequation 10x + 8y ≤ 240. So,the region which contains the origin represents the solution set of the inequation 10x + 8y ≤ 240.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + 8y ≤ 200, 10x + 8y ≤ 240, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are O(0, 0), B_{1}(0, 25), E_{1}(8, 20), C_{1}(24, 0).
The values of Z at these corner points are as follows.
Corner point  Z = 0.5x + 0.6y 
O(0, 0)  0 
B_{1}(0, 25)  15 
E_{1}(8, 20)  16 
C_{1}(24, 0)  12 
The maximum value of Z is 160 at E_{1}(8, 20).
Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs 16.
Page No 30.53:
Question 19:
A manufacturer makes two products A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hours of production and weekly turnover must not be less than Rs 10000. If the profit on each of product A is Rs 20 and on product B is Rs 30, then how many of each should be produced so that the profit is maximum. Also, find the maximum profit.
Answer:
Let x units of product A and y units of product B were manufactured.
Number of units cannot be negative.
Therefore, $x,y\ge 0$
According to question, the given information can be tabulated as:
Selling price(Rs)  Manufacturing time(hrs)  
Product A(x)  200  0.5 
Product B(y)  300  1 
Also, the availability of time is 40 hours and the revenue should be atleast Rs 10000.
Further, it is given that there is a permanent order for 14 units of product A and 16 units of product B.
Therefore, the constraints are
$200x+300y\ge 10000\phantom{\rule{0ex}{0ex}}0.5x+y\le 40\phantom{\rule{0ex}{0ex}}x\ge 14\phantom{\rule{0ex}{0ex}}y\ge 16$
If the profit on each of product A is Rs 20 and on product B is Rs 30.Therefore, profit gained on x units of product A and y units of product B is Rs 20x and Rs 30y respectively.
Total profit = Z = $20x+30y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $20x+30y$
subject to
$2x+3y\ge 100\phantom{\rule{0ex}{0ex}}x+2y\le 80\phantom{\rule{0ex}{0ex}}x\ge 14\phantom{\rule{0ex}{0ex}}y\ge 16$
$x,y\ge 0$
First we will convert inequations into equations as follows:
2x + 3y = 100, x + 2y = 80, x = 14, y = 16, x = 0 and y = 0.
Region represented by 2x + 3y ≥ 100:
The line 2x + 3y = 100 meets the coordinate axes at A_{1}(50, 0) and ${B}_{1}\left(0,\frac{100}{3}\right)$ respectively. By joining these points we obtain the line
2x + 3y = 100 . Clearly (0,0) does not satisfies the 2x + 3y = 100. So, the region which does not contain the origin represents the solution set of the inequation 2x + 3y ≥ 100.
Region represented by x + 2y ≤ 80:
The line x + 2y = 80 meets the coordinate axes at C_{1}(80, 0) and D_{1}(0, 40) respectively. By joining these points we obtain the line
x + 2y = 80. Clearly (0,0) satisfies the inequation x + 2y ≤ 80. So,the region which contains the origin represents the solution set of the inequation x + 2y ≤ 80.
Region represented by x ≥ 14
x = 14 is the line passes through (14, 0) and is parallel to the Y axis.The region to the right of the line x = 14 will satisfy the inequation.
Region represented by y ≥ 16
y = 16 is the line passes through (0, 16) and is parallel to the X axis.The region above the line y = 16 will satisfy the inequation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 3y ≥ 100, x + 2y ≤ 80, x ≥ 14, y ≥ 16, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E_{1}(26, 16), F_{1}(48, 16), G_{1}(14, 33) and H_{1}(14, 24)
The values of Z at these corner points are as follows
Corner point  Z= 20x + 30y 
E_{1}  1000 
F_{1}  1440 
G_{1}  1270 
H_{1}  1000 
The maximum value of Z is Rs 1440 which is attained at F_{1}$\left(48,16\right)$.
Thus, the maximum profit is Rs 1440 obtained when 48 units of product A and 16 units of product B were manufactured.
Page No 30.53:
Question 20:
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?
Answer:
Let x trunks of first type and y trunks of second type were manufactured.
Number of trunks cannot be negative.Therefore,
$x,y\ge 0$
According to question, the given information can be tabulated as
Machine A(hrs)  Machine B(hrs)  
First type(x)  3  3 
Second type(y)  3  2 
Availability  18  15 
Therefore, the constraints are
$3x+3y\le 18\phantom{\rule{0ex}{0ex}}3x+2y\le 15$
He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively.Therefore, profit gained by him from x trunks of first type and y trunks of second type is Rs 30x and Rs 25y respectively.
Total profit = Z = $30x+25y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $30x+25y$
subject to
$3x+3y\le 18\phantom{\rule{0ex}{0ex}}3x+2y\le 15$
$x,y\ge 0$
First we will convert inequations into equations as follows:
3x + 3y = 18, 3x + 2y = 15, x = 0 and y = 0
Region represented by 3x + 3y ≤ 18:
The line 3x + 3y = 18 meets the coordinate axes at A_{1}(6, 0) and B_{1}(0, 6) respectively. By joining these points we obtain the line 3x + 3y = 18. Clearly (0,0) satisfies the 3x + 3y = 18. So, the region which contains the origin represents the solution set of the inequation 3x + 3y ≤ 18.
Region represented by 3x + 2y ≤ 15:
The line 3x + 2y = 15 meets the coordinate axes at C_{1}(5, 0) and ${D}_{1}\left(0,\frac{15}{2}\right)$ respectively. By joining these points we obtain the line 3x + 2y = 15. Clearly (0,0) satisfies the inequation 3x + 2y ≤ 15. So,the region which contains the origin represents the solution set of the inequation 3x + 2y ≤ 15.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 3x + 3y ≤ 18, 3x + 2y ≤ 15, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), B_{1}(0, 6), E_{1}(3, 3) and C_{1}(5, 0).
The values of Z at these corner points are as follows
Corner point  Z = 30x + 25y 
O  0 
B_{1}  150 
E_{1}  165 
C_{1}  150 
The maximum value of Z is 165 which is attained at E_{1}(3, 3).
Thus, the maximum profit is Rs 165 obtained when 3 units of each type of trunk is manufactured.
Page No 30.53:
Question 21:
A manufacturer of patent medicines is preparing a production plan on medicines, A and B. There are sufficient raw materials available to make 20000 bottles of A and 40000 bottles of B, but there are only 45000 bottles into which either of the medicines can be put. Further, it takes 3 hours to prepare enough material to fill 1000 bottles of A, it takes 1 hour to prepare enough material to fill 1000 bottles of B and there are 66 hours available for this operation. The profit is Rs 8 per bottle for A and Rs 7 per bottle for B. How should the manufacturer schedule his production in order to maximize his profit?
Answer:
Let x bottles of medicine A and y bottles of medicine B are prepared.
Number of bottles cannot be negative.
Therefore, $x,y\ge 0$
According to question, the constraints are
$x\le 20000\phantom{\rule{0ex}{0ex}}y\le 40000\phantom{\rule{0ex}{0ex}}x+y\le 45000\phantom{\rule{0ex}{0ex}}$
It takes 3 hours to prepare enough material to fill 1000 bottles of A, it takes 1 hour to prepare enough material to fill 1000 bottles of B
Time taken to fill one bottle of A is $\frac{3}{1000}$hrs and time taken by to fill one bottle of B is $\frac{1}{1000}$hrs.
Therefore, time taken to fill x bottles of A and y bottles of B is $\frac{3x}{1000}$hrs and $\frac{y}{1000}$hrs respectively.
It is given that there are 66 hours available for this operation.
$\therefore \frac{3x}{1000}+\frac{y}{1000}\le 66$
The profit is Rs 8 per bottle for A and Rs 7 per bottle for B.Therefore, profit gained on x bottles of medicine A and y bottles of medicine B is 8x and 7y respectively.
Total profit = Z = $8x+7y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $8x+7y$
subject to
$x\le 20000\phantom{\rule{0ex}{0ex}}y\le 40000\phantom{\rule{0ex}{0ex}}x+y\le 45000\phantom{\rule{0ex}{0ex}}$
$\frac{3x}{1000}+\frac{y}{1000}\le 66\Rightarrow 3x+y\le 66000$
$x,y\ge 0$
First we will convert inequations into equations as follows:
x =20000, y = 40000, x + y = 45000, 3x + y = 66000, x = 0 and y = 0
Region represented by x ≤ 20000:
The line x = 20000 is the line that passes through A_{1}(20000, 0) and is parallel to Y axis.The region to the left of the line x = 20000 will satisfy the inequation x ≤ 20000.
Region represented by y ≤ 40000:
The line y = 40000 is the line that passes through B_{1}(0, 40000) and is parallel to X axis.The region below the line y = 40000 will satisfy the inequation y ≤ 40000.
Region represented by x + y ≤ 45000:
The line x + y = 45000 meets the coordinate axes at C_{1}(45000, 0) and D_{1}(0, 45000) respectively. By joining these points we obtain the line x + y = 45000. Clearly (0,0) satisfies the inequation x + y ≤ 45000. So,the region which contains the origin represents the solution set of the inequation x + y ≤ 45000.
Region represented by 3x + y ≤ 66000:
The line 3x + y = 66000 meets the coordinate axes at E_{1}(22000, 0) and ${F}_{1}\left(0,66000\right)$ respectively. By joining these points we obtain the line 3x + y = 66000. Clearly (0,0) satisfies the inequation 3x + y ≤ 66000. So,the region which contains the origin represents the solution set of the inequation 3x + y ≤ 66000.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x ≤ 20000, y ≤ 40000, x + y ≤ 45000, 3x + y ≤ 66000, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), B_{1}(0, 40000), G_{1}(10500, 34500), H_{1}(6000, 20000) and A_{1}(20000, 0).
The values of Z at these corner points are as follows
Corner point  Z= 8x + 7y 
O  0 
B_{1}  280000 
G_{1}  325500 
H_{1}  188000 
A_{1}  160000 
The maximum value of Z is 325500 which is attained at G_{1}(10500, 34500).
Thus, the maximum profit is Rs 325500 obtained when 10500 bottles of A and 34500 bottles of B were manufactured.
Page No 30.53:
Question 22:
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 400 is made on each first class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to the first class. Determine how many each type of tickets must be sold in order to maximize the profit for the airline. What is the maximum profit.
Answer:
Let x tickets of first class and y tickets of economy class were sold.
Number of tickets cannot be negative.
Therefore, $x,y\ge 0$
An aeroplane can carry a maximum of 200 passengers.
$x+y\le 200$
The airline reserves at least 20 seats of first class and at least 4 times as many passengers prefer to travel by economy class to the first class.
$x\ge 20\phantom{\rule{0ex}{0ex}}y\ge 4x\phantom{\rule{0ex}{0ex}}$
A profit of Rs 400 is made on each first class ticket and a profit of Rs 600 is made on each economy class ticket.Therefore, profit gained by selling x tickets of first class and y tickets of economy class is Rs 400x and Rs 600y respectively.
Total profit = Z = $400x+600y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $400x+600y$
subject to
$x+y\le 200$
$x\ge 20\phantom{\rule{0ex}{0ex}}y\ge 4x\phantom{\rule{0ex}{0ex}}$
$x,y\ge 0$
First we will convert inequations into equations as follows:
x + y =200, x = 20, y = 4x, x = 0 and y = 0
Region represented by x + y ≤ 200:
The line x + y = 200 meets the coordinate axes at A(200, 0) and B(0, 200) respectively. By joining these points we obtain the line
x + y = 200. Clearly (0,0) satisfies the inequation x + y ≤ 200. So,the region which contains the origin represents the solution set of the inequation x + y ≤ 200.
Region represented by x ≥ 20:
The line x = 20 is the line that passes through (20, 0) and is parallel to Y axis.The region to the right of the line x = 20 will satisfy the inequation x ≥ 20.
Region represented by y ≥ 4x:
The line y = 4x is the line that passes through (0, 0).The region above the line y = 4x will satisfy the inequation y ≥ 4x.Let us understand by taking one point which is below the line y = 4x. Let it be (1, 3).Here, 3 < 4 which does not satisfy the inequation y ≥ 4x.
Hence, the region above the line y = 4x will satisfy the inequation y ≥ 4x.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 200, x ≥ 20, y ≥ 4x, x ≥ 0 and y ≥ 0 are as follows.
The corner points are C(20, 80), D(40, 160), E(20, 180).
The values of Z at these corner points are as follows
Corner point  Z = 400x + 600y 
C  56000 
D  112000 
E  116000 
The maximum value of Z is 116000 which is attained at C(20, 180).
Thus, the maximum profit is Rs 116000 obtained when 20 first class tickets and 180 economy class tickets were sold
Page No 30.53:
Question 23:
A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Answer:
Let x kg of type I fertilizer and y kg of type II fertilizer are supplied.
Quantity of fertilizer cannot be negative.
Therefore, $x,y\ge 0$
A gardener has supply of fertilizer of type I which consists of 10% nitrogen and type II fertilizer consists of 5% nitrogen and he needs at least 14 kg of nitrogen for his crop.
$\frac{10x}{100}+\frac{5x}{100}\ge 14\Rightarrow 10x+5x\ge 1400$
A gardener has supply of fertilizer of type I which consists 6% phosphoric acid and type II fertilizer consists of 10% phosphoric acid.
And he needs 14 kg of phosphoric acid for his crop.
$\frac{6x}{100}+\frac{10x}{100}\ge 14\Rightarrow 6x+10x\ge 1400$
Therefore, according to the question, constraints are
$10x+5y\ge 1400\phantom{\rule{0ex}{0ex}}6x+10y\ge 1400$
If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg.Therefore, cost of x kg of type I fertilizer and y kg of type II fertilizer is Rs 0.60x and Rs 0.40y respectively.
Total cost = Z = $0.60x+0.40y$ which is to be minimised.
Thus, the mathematical formulation of the given linear programmimg problem is
Min Z = $0.60x+0.40y$
subject to
$6x+10y\ge 1400\phantom{\rule{0ex}{0ex}}10x+5y\ge 1400$
$x,y\ge 0$
First we will convert inequations into equations as follows:
6x + 10y = 1400, 10x + 5y = 1400, x = 0 and y = 0
Region represented by 6x + 10y ≥ 1400:
The line 6x + 10y = 1400 meets the coordinate axes at $A\left(\frac{700}{3},0\right)$ and B(0, 140) respectively. By joining these points we obtain the line
6x + 10y = 1400. Clearly (0,0) does not satisfies the 6x + 10y = 1400. So, the region which does not contain the origin represents the solution set of the inequation 6x + 10y ≥1400.
Region represented by 10x + 5y ≥ 1400:
The line 10x + 5y = 1400 meets the coordinate axes at C(140, 0) and $D\left(0,280\right)$ respectively. By joining these points we obtain the line
10x + 5y = 1400. Clearly (0,0) does not satisfies the inequation 10x + 5y ≥ 1400. So,the region which does not contain the origin represents the solution set of the inequation 10x + 5y ≥ 1400.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 6x + 10y ≥1400, 10x + 5y ≥ 1400, x ≥ 0 and y ≥ 0 are as follows.
The corner points are D(0, 280), E(100, 80) and $A\left(\frac{700}{3},0\right)$.
The values of Z at these corner points are as follows
Corner point  Z= 0.60x + 0.40y 
D  112 
E  92 
A  140 
The minimum value of Z is Rs 92 which is attained at E(100, 80).
Thus, the minimum cost is Rs 92 obtained when 100 kg of type I fertilizer and 80 kg of type II fertilizer were supplied.
Page No 30.53:
Question 24:
Anil wants to invest at most Rs 12000 in Saving Certificates and National Saving Bonds. According to rules, he has to invest at least Rs 2000 in Saving Certificates and at least Rs 4000 in National Saving Bonds. If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum, how much money should he invest to earn maximum yearly income? Find also his maximum yearly income.
Answer:
Let Anil invests Rs x in Saving certificates and Rs y in National Saving bonds.
Therefore,
$x,y\ge 0$
Anil wants to invest at most Rs 12000 in Saving Certificates and National Saving Bonds.
$x+y\le 12000\phantom{\rule{0ex}{0ex}}$
According to rules, he has to invest at least Rs 2000 in Saving Certificates and at least Rs 4000 in National Saving Bonds.
$x\ge 2000\phantom{\rule{0ex}{0ex}}y\ge 4000$
If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum.
Total earning from investment = Z = $\frac{8x}{100}+\frac{10y}{100}$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $\frac{8x}{100}+\frac{10y}{100}$
subject to
$x+y\le 12000\phantom{\rule{0ex}{0ex}}x\ge 2000\phantom{\rule{0ex}{0ex}}y\ge 4000$
$x,y\ge 0$
First we will convert inequations into equations as follows:
x + y =12000, x = 2000, y = 4000, x = 0 and y = 0
Region represented by x + y ≤ 12000:
The line x + y = 12000 meets the coordinate axes at A(12000, 0) and B(0, 12000) respectively. By joining these points we obtain the line
x + y = 12000. Clearly (0,0) satisfies the inequation x + y ≤ 12000. So,the region which contains the origin represents the solution set of the inequation x + y ≤ 12000.
Region represented by x ≥ 2000:
The line x = 2000 is the line that passes through (2000, 0) and is parallel to Y axis.The region to the right of the line x = 2000 will satisfy the inequation x ≥ 2000.
Region represented by y ≥ 4000:
The line y = 4000 is the line that passes through (0, 4000) and is parallel to X axis.The region above the line y = 4000 will satisfy the inequation y ≥ 4000.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints is
The corner points are E(2000, 10000), D(8000, 4000), C(2000, 4000)
The values of Z at these corner points are as follows
Corner point  Z= $\frac{8x}{100}+\frac{10y}{100}$ 
E  1160 
D  1040 
C  560 
The maximum value of Z is 1160 which is attained at E(2000, 10000.
Thus, the maximum earning is Rs 1160 obtained when Rs 2000 were invested in Saving's certificates and Rs 10000 were invested in National Saving Bond.
Page No 30.53:
Question 25:
A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type A produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?
Answer:
Let the man planted x trees of type A and y trees of type B.
Number of trees cannot be negative.
Therefore, $x,y\ge 0$
To plant tree of type A requires 10 sq.m and type B requires 20 sq.m of ground per tree. And, it is given that a man owns a field of area 1000 sq.m.Therefore,
$10x+20y\le 1000\phantom{\rule{0ex}{0ex}}$
Type A costs Rs 20 per tree and type B costs Rs 25 per tree.Therefore, x trees of type A and y trees of type B costs Rs 20x and Rs 25y respectively. A man has a sum of Rs 1400 to purchase young trees.
$20x+25y\le 1400$
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = 40x − 20x + 60y − 25y = 20x + 35y
subject to
$10x+20y\le 1000\phantom{\rule{0ex}{0ex}}20x+25y\le 1400$
The feasible region determined by the system of constraints is
The corner points are A(0, 50), B(20, 40), C(70, 0)
The values of Z at these corner points are as follows
Corner point  Z = 20x + 35y 
A  1750 
B  1800 
C  1400 
The maximum value of Z is 1800 which is attained at B(20, 40)
Thus, the maximum profit is Rs 1800 obtained when Rs 20 were invested on type A and Rs 40 were invested on type B.
Page No 30.53:
Question 26:
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is ₹5.00 and a shade is ₹3.00. Assuming that the manufacturer sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? [NCERT, CBSE 2013]
Answer:
Suppose x units of pedestal lamps and y units of wooden shades are produced on a day to maximise the profit.
Since a pedestal lamp requires 2 hours on the grinding/cutting machine and a wooden shade requires 1 hour on the grinding/cutting machine, therefore, the total hours required for grinding/cutting x units of pedestal lamps and y units of wooden shades are (2x + y). But, the grinding/cutting machine is available for at most 12 hours on a day.
∴ 2x + y ≤ 12
Similarly, a pedestal lamp requires 3 hours on the sprayer and a wooden shade requires 2 hours on the sprayer, therefore, the total hours required for spraying x units of pedestal lamps and y units of wooden shades are (3x + 2y). But, the sprayer is available for at most 20 hours on a day.
∴ 3x + 2y ≤ 20
The profit from the sale of a pedestal lamp is ₹5.00 and a wooden shade is ₹3.00. Therefore, the total profit from the sale of x units of pedestal lamps and y units of wooden shades is ₹(5x + 3y).
Thus, the given linear programming problem is
Maximise Z = 5x + 3y
subject to the constraints
2x + y ≤ 12
3x + 2y ≤ 20
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(6, 0), B(4, 4) and C(0, 10).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 5x + 3y 
(0, 0)  5 × 0 + 3 × 0 = 0 
(6, 0)  5 × 6 + 3 × 0 = 30 
(4, 4)  5 × 4 + 3 × 4 = 32 → Maximum 
(0, 10)  5 × 0 + 3 × 10 = 30 
The maximum value of Z is 32 at x = 4, y = 4.
Hence, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profit. The maximum profit of the manufacturer is ₹32 on a day.
Page No 30.54:
Question 27:
A producer has 30 and 17 units of labour and capital respectively which he can use to produce two type of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital is required to produce one unit of y. If x and y are priced at Rs 100 and Rs 120 per unit respectively, how should be producer use his resources to maximize the total revenue? Solve the problem graphically.
Answer:
Let ${x}_{1}$ and ${y}_{1}$ units of goods x and y were produced respectively.
Number of units of goods cannot be negative.
Therefore, ${x}_{1},{y}_{1}\ge 0$
To produce one unit of x, 2 units of labour and for one unit of y, 3 units of labour are required.
$2{x}_{1}+3{y}_{1}\le 30\phantom{\rule{0ex}{0ex}}$
To produce one unit of x, 3 units of capital is required and 1 unit of capital is required to produce one unit of y
$3{x}_{1}+{y}_{1}\le 17$
If x and y are priced at Rs 100 and Rs 120 per unit respectively, Therefore, cost of x_{1} and y_{1} units of goods x and y is Rs 100x_{1} and Rs 120y_{1} respectively.
Total revenue = Z = $100{x}_{1}+120{y}_{1}$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $100{x}_{1}+120{y}_{1}$
subject to
$2{x}_{1}+3{y}_{1}\le 30\phantom{\rule{0ex}{0ex}}3{x}_{1}+{y}_{1}\le 17$
$x,y\ge 0$
First we will convert inequations into equations as follows:
2x_{1} + 3y_{1} = 30, 3x_{1} + y_{1} = 17, x = 0 and y = 0
Region represented by 2x_{1} + 3y_{1} ≤ 30:
The line 2x_{1} + 3y_{1}_{ }= 30 meets the coordinate axes at A(15, 0) and B(0, 10) respectively. By joining these points we obtain the line
2x_{1} + 3y_{1}_{ }= 30. Clearly (0,0) satisfies the 2x_{1} + 3y_{1}_{ }= 30. So, the region which contains the origin represents the solution set of the inequation 2x_{1} + 3y_{1} ≤ 30.
Region represented by 3x_{1} + y_{1} ≤ 17:
The line 3x_{1} + y_{1} = 17 meets the coordinate axes at $C\left(\frac{17}{3},0\right)$ and $D\left(0,17\right)$ respectively. By joining these points we obtain the line
3x_{1} + y_{1} = 17. Clearly (0,0) satisfies the inequation 3x_{1} + y_{1} ≤ 17. So,the region which contains the origin represents the solution set of the inequation 3x_{1} + y_{1} ≤ 17.
Region represented by x_{1} ≥ 0 and y_{1} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x_{1} + 3y_{1} ≤ 30, 3x_{1} + y_{1} ≤ 17, x ≥ 0 and y ≥ 0 are as follows.
The corner points are B(0, 10), E(3, 8) and $C\left(\frac{17}{3},0\right)$ .
The values of Z at these corner points are as follows
Corner point  Z= $100{x}_{1}+120{y}_{1}$ 
B  1200 
E  1260 
C  $\frac{1700}{3}$ 
The maximum value of Z is 1260 which is attained at E(3, 8).
Thus, the maximum revenue is Rs 1260 obtained when 3 units of x and 8 units of y were produced.
Page No 30.54:
Question 28:
A firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B. Each product is processed on two machines M_{1} and M_{2}. One unit of type A requires one minute of processing time on M_{1} and two minutes of processing time on M_{2}, whereas one unit of type B requires one minute of processing time on M_{1} and one minute on M_{2}. Machines M_{1} and M_{2} are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.
Answer:
Let x units of product A and y units of product B were manufactured.
Number of products cannot be negative.
Therefore, $x,y\ge 0$
According to question, the given information can be tabulated as
Time on ${\mathrm{M}}_{1}$(minutes)  Time on ${\mathrm{M}}_{2}$(minutes)  
Product A(x)  1  2 
Product B(y)  1  1 
Availability  300  360 
The constraints are
$x+y\le 300$, $2x+y\le 360$
Firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B.Therefore, x units of product A and y units of product B costs Rs 5x and Rs 3y respectively.
Total profit = Z = $5x+3y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $5x+3y$
subject to
$x+y\le 300$,
$2x+y\le 360$
$x,y\ge 0$
First we will convert inequations into equations as follows:
x + y = 300, 2x + y = 360, x = 0 and y = 0
Region represented by x + y ≤ 300:
The line x + y = 300 meets the coordinate axes at A_{1}(300, 0) and B_{1}(0, 300) respectively. By joining these points we obtain the line
x + y = 30. Clearly (0,0) satisfies the x + y = 30. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 300.
Region represented by 2x + y ≤ 360:
The line 2x + y = 360 meets the coordinate axes at C_{1}(180, 0) and D_{1}(0, 360) respectively. By joining these points we obtain the line
2x + y = 360. Clearly (0,0) satisfies the inequation 2x + y ≤ 360. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 360.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 300, 2x + y ≤ 360, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), B_{1}(0, 300), E_{1}(60, 240) and C_{1}(180, 0).
The values of Z at these corner points are as follows
Corner point  Z= 5x + 3y 
O  0 
B_{1}  900 
E_{1}  1020 
C_{1}  900 
The maximum value of Z is Rs 1020 which is attained at B_{1}$\left(60,240\right)$.
Thus, the maximum profit is Rs 1020 obtained when 60 units of product A and 240 units of product B were manufactured.
Page No 30.54:
Question 29:
A small firm manufacturers items A and B. The total number of items A and B that it can manufacture in a day is at the most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.
Answer:
Let the firm manufacture x items of A and y items of B per day.
Number of items cannot be negative.
Therefore, x ≥0 and y ≥0.
It is given that the total number of items manufactured per day is at most 24.
∴ x + y ≤ 24
Item A takes 1 hour to make and item B takes 0.5 hour to make.
The maximum number of hours available per day is 16 hours.
∴ x + 0.5y ≤ 16
If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160.Therefore, profit gained on x items of A and y items of B is
Rs 300x and Rs 160y respectively.
∴ Total profit, Z = 300x + 160y
Thus, the mathematical formulation of the given problem is:
Maximise Z = 300x + 160y
subject to the constraints
x+ y ≤ 24
x + 0.5y ≤ 16
x, y ≥0
First we will convert inequations into equations as follows:
x + y = 24, x + 0.5y = 16, x = 0 and y = 0
Region represented by x + y ≤ 24:
The line x + y = 24 meets the coordinate axes at A_{1}(24, 0) and B_{1}(0, 24) respectively. By joining these points we obtain the line x + y = 24. Clearly (0,0) satisfies the x + y = 24. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 24.
Region represented by x + 0.5y ≤ 16:
The line x + 0.5y = 16 meets the coordinate axes at C_{1}(16, 0) and D_{1}(0, 32) respectively. By joining these points we obtain the line x + 0.5y = 16. Clearly (0,0) satisfies the inequation x + 0.5y ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + 0.5y ≤ 16.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0) ,C_{1}(16, 0), E_{1}(8, 16) and B_{1}(0, 24).
The value of Z at these corner points are as follows:
Corner point  Z = 300x + 160y 
O(0, 0)  0 
C_{1}(16, 0)  4800 
E_{1}(8, 16)  4960 
B_{1}(0, 24)  3840 
Thus, the maximum value of Z is 4960 at E_{1}(8, 16).
Thus, 8 units of item A and 16 units of item B should be manufactured per day to maximise the profits.
Page No 30.54:
Question 30:
A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
Answer:
Let x toys of type A and y toys of type B were manufactured.
The given information can be tabulated as follows:
Cutting time (minutes)  Assembling time (minutes)  
Toy A(x)  5  10 
Toy B(y)  8  8 
Availability  180  240 
The constraints are
$5x+8y\le 180\phantom{\rule{0ex}{0ex}}10x+8y\le 240$
The profit is Rs 50 each on type A and Rs 60 each on type B. Therefore, profit gained on x toys of type A and y toys of type B is Rs 50x and Rs 60 y respectively.
Total profit = Z = $50x+60y$
The mathematical formulation of the given problem is
Max Z = $50x+60y$
subject to
$5x+8y\le 180\phantom{\rule{0ex}{0ex}}10x+8y\le 240$
First we will convert inequations into equations as follows:
5x + 8y = 180, 10x + 8y = 240, x = 0 and y = 0
Region represented by 5x + 8y ≤ 180:
The line 5x + 8y = 180 meets the coordinate axes at A_{1}(36, 0) and ${B}_{1}\left(0,\frac{45}{2}\right)$ respectively. By joining these points we obtain the line 5x + 8y = 180. Clearly, (0,0) satisfies the 5x + 8y = 180. So, the region which contains the origin represents the solution set of the inequation 5x + 8y ≤ 180.
Region represented by 10x + 8y ≤ 240:
The line 10x + 8y = 240 meets the coordinate axes at C_{1}(24, 0) and ${D}_{1}\left(0,30\right)$ respectively. By joining these points we obtain the line 10x + 8y = 240. Clearly (0,0) satisfies the inequation 10x + 8y ≤ 240. So,the region which contains the origin represents the solution set of the inequation 10x + 8y ≤ 240.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + 8y ≤ 180, 10x + 8y ≤ 240, x ≥ 0 and y ≥ 0 are as follows.
The feasible region is shown in the figure
The corner points are B_{1}$\left(0,\frac{45}{2}\right)$, E_{1}(12, 15) and C_{1}(24, 0).
The values of Z at the corner points are
Corner points  Z = $50x+60y$ 
O  0 
B_{1}  1350 
E_{1}  1500 
C_{1}  1200 
Thus, for maximum profit, 12 units of toy A and 15 units of toy B should be manufactured.
Page No 30.54:
Question 31:
Answer:
Let $x$ units and y units of articles A and B are produced respectively.
Number of articles cannot be negative.
Therefore, $x,y\ge 0$
The product of each unit of article A requires 4 hours in assembly and that of article B requires 2 hours in assembly and the maximum capacity of the assembly department is 60 hours a week
$4x+2y\le 60$
The product of each unit of article A requires 2 hours in finishing and that of article B requires 4 hours in assembly and the maximum capacity of the finishing department is 48 hours a week.
$2x+4y\le 48\phantom{\rule{0ex}{0ex}}$
If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B. Therefore, profit gained from $x$ units and y units of articles A and B respectively is Rs 6x and Rs 8y respectively.
Total revenue = Z = $6x+8y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $6x+8y$
subject to
$2x+4y\le 48\phantom{\rule{0ex}{0ex}}4x+2y\le 60$
$x,y\ge 0$
First we will convert inequations into equations as follows:
2x + 4y = 48, 4x + 2y = 60, x = 0 and y = 0
Region represented by 2x + 4y ≤ 48:
The line 2x + 4y = 48 meets the coordinate axes at A_{1}(24, 0) and ${B}_{1}\left(0,12\right)$ respectively. By joining these points we obtain the line 2x + 4y = 48. Clearly (0,0) satisfies the 2x + 4y = 48. So, the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 48.
Region represented by 4x + 2y ≤ 60:
The line 4x + 2y = 60 meets the coordinate axes at C_{1}(15, 0) and ${D}_{1}\left(0,30\right)$ respectively. By joining these points we obtain the line 4x + 2y = 60. Clearly (0,0) satisfies the inequation 4x + 2y ≤ 60. So,the region which contains the origin represents the solution set of the inequation 4x + 2y ≤ 60.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 4y ≤ 48, 4x + 2y ≤ 60, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), B_{1}(0, 12), E_{1}(12, 6) and C_{1}(15,0).
The values of Z at these corner points are as follows
Corner point  Z= 6x + 8y 
O  0 
B_{1}  96 
E_{1}  120 
C_{1}  90 
The maximum value of Z is 120 which is attained at E_{1}(12, 6).
Thus, the maximum profit is Rs 120 obtained when 12 units of article A and 6 units of article B were manufactured.
Page No 30.54:
Question 32:
A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs 300 and on one item of B is Rs 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
Answer:
Let x and y be the number of items of A and B that should be produced each day to maximize the profit.
Number of items cannot be negative.
Therefore,
It is also given that the firm can produce at most 24 items in a day.
∴ x+ y ≤ 24
Also, the time required to make an item of A is one hour and time required to make an item of B is half an hour.
Therefore, the time required to produce x items of A and y items of B is hours. However, the maximum time available in a day is 16 hours.
It is given that the profit on an item of A is Rs 300 and on one item of B is Rs 160. Therefore, the profit gained from x items of A and y items of B is Rs 300x and Rs 160y respectively.
Total profit Z = 300x + 160y
The mathematical form of the given LPP is:
Maximize Z = 300x + 160y
subject to constraints:
$x+y\le 24\phantom{\rule{0ex}{0ex}}x+\frac{1}{2}y\le 16\phantom{\rule{0ex}{0ex}}x\ge 0,y\ge 0$
First we will convert inequations into equations as follows:
x + y = 24, x + $\frac{1}{2}$y = 16, x = 0 and y = 0
Region represented by x + y ≤ 24:
The line x + y = 24 meets the coordinate axes at A_{1}(24, 0) and ${B}_{1}\left(0,24\right)$ respectively. By joining these points we obtain the line x + y = 24. Clearly, (0,0) satisfies the x + y = 24. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 24.
Region represented by x + $\frac{1}{2}$y ≤ 16:
The line x + $\frac{1}{2}$y = 16 meets the coordinate axes at C_{1}(16, 0) and ${D}_{1}\left(0,32\right)$ respectively. By joining these points we obtain the line x + $\frac{1}{2}$y = 16. Clearly, (0,0) satisfies the inequation x + $\frac{1}{2}$y ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + $\frac{1}{2}$y ≤ 16.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 24, x + $\frac{1}{2}$y ≤ 16, x ≥ 0 and y ≥ 0 are as follows.
The feasible region is shown in the figure
GRAPH
In the above graph, the shaded region is the feasible region. The corner points are O(0, 0), C_{1}(16, 0), E_{1}(8, 16), and B_{1}(0, 24).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points 
Z = 300x + 160y 

O(0, 0) 
0 

C_{1}(16, 0) 
4800 

E_{1}(8, 16) 
4960 
Maximum 
B_{1}(0, 24) 
3840 
Clearly, Z is maximum at x = 8 and y = 16 and the maximum value of Z at this point is 4960.
Thus, 8 items of A and 16 items of B should be produced in order to maximize the profit and the maximum profit is Rs 4960.
Page No 30.54:
Question 33:
A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 manhours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B, how many units of each product should be sold to maximize profit?
Answer:
Let x units of product A and y units of product B were manufactured.
Clearly, $x\ge 0,y\ge 0$
It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.The two products are produced in a common production process, which has a total capacity of 500 manhours.
$5x+3y\le 500\phantom{\rule{0ex}{0ex}}$
The maximum number of unit of A that can be sold is 70 and that for B is 125.
$x\le 70\phantom{\rule{0ex}{0ex}}y\le 125$
If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B. Therefore, profit x units of product A and y units of product B is Rs 20x and Rs 15y respectively.
Total profit = Z = $20x+15y$
The mathematical formulation of the given problem is
Max Z = $20x+15y$
subject to
$5x+3y\le 500\phantom{\rule{0ex}{0ex}}x\le 70\phantom{\rule{0ex}{0ex}}y\le 125$
$x\ge 0,y\ge 0$
First we will convert inequations into equations as follows:
5x + 3y = 500, x = 70, y = 125, x = 0 and y = 0
Region represented by 5x + 3y ≤ 500:
The line 5x + 3y = 500 meets the coordinate axes at A_{1}(100, 0) and ${B}_{1}\left(0,\frac{500}{3}\right)$ respectively. By joining these points we obtain the line 5x + 3y = 500. Clearly (0,0) satisfies the 5x + 3y = 500. So, the region which contains the origin represents the solution set of the inequation 5x + 3y ≤ 500.
Region represented by x ≤ 70:
The line x = 70 is the line passes through C_{1}(70, 0) and is parallel to Y axis. The region to the left of the line x = 70 will satisfy the inequation x ≤ 70.
Region represented by y ≤ 125:
The line y = 125 is the line passes through D_{1}(0, 125) and is parallel to X axis. The region below the the line y = 125 will satisfy the inequation y ≤ 125.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + 3y ≤ 500, x ≤ 70, y ≤ 125, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), D_{1}$\left(0,125\right)$, E_{1}(25, 125), F_{1}(70, 50) and C_{1}(70, 0).
The values of Z at the corner points are
Corner points  Z = $20x+15y$ 
O  0 
D_{1}  1875 
E_{1}  2375 
F_{1}  2150 
C_{1}  1400 
Thus, maximum profit is Rs 2375, 25 units of A and 125 units of B should be manufactured.
Page No 30.54:
Question 34:
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs 3 and Rs 2 per box, how many of each should be made in order to maximize the total profit?
Answer:
Let x large boxes and y small boxes be manufactured.
Number of boxes cannot be negative.
Therefore, $x\ge 0,y\ge 0$
The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box and if 60 sq. metre of cardboard is in stock.
$4x+3y\le 60\phantom{\rule{0ex}{0ex}}$
The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes.
$x\ge 3\phantom{\rule{0ex}{0ex}}y\ge 2x$
If the profits on the large and small boxes are Rs 3 and Rs 2 per box. Therefore, profit gained by him on x large boxes and y small boxes is Rs 3x and Rs 2y respectively.
Total profit = Z = $3x+2y$
The mathematical formulation of the given problem is
Max Z = $3x+2y$
subject to
$4x+3y\le 60\phantom{\rule{0ex}{0ex}}x\ge 3\phantom{\rule{0ex}{0ex}}y\ge 2x$
$x\ge 0,y\ge 0$
First we will convert inequations into equations as follows:
4x + 3y = 60, x = 3, y = 2x, x = 0 and y = 0
Region represented by 4x + 3y ≤ 60:
The line 4x + 3y = 60 meets the coordinate axes at A(15, 0) and B(0, 20) respectively. By joining these points we obtain the line
4x + 3y = 60. Clearly (0,0) satisfies the 4x + 3y = 60. So, the region which contains the origin represents the solution set of the inequation 4x + 3y ≤ 60.
Region represented by x ≥ 3:
The line x = 3 is the line passes through (3, 0) and is parallel to Y axis. The region to the right of the line x = 3 will satisfy the inequation
x ≥ 3.
Region represented by y ≥ 2x:
The line y = 2x is the line that passes through (0, 0). The region above the line y = 2x will satisfy the inequation y ≥ 2x. Like if we take an example taking a point (5, 1) below the line y = 2x .Here, 1 < 10 which does not satisfies the inequation y ≥ 2x. Hence, the region above the line y = 2x will satisfy the inequation y ≥ 2x.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 4x + 3y ≤ 60, x ≥ 3, y ≥ 2x, x ≥ 0 and y ≥ 0 are as follows
The corner points are E$\left(3,16\right)$, D(6, 12) and C(3, 6).
The values of Z at the corner points are
Corner points  Z = $3x+2y$ 
E  41 
D  42 
C  21 
Thus, for maximum profit is Rs 42, 6 units of large boxes and 12 units of smaller boxes should be manufactured.
Page No 30.54:
Question 35:
A manufacturer makes two products, A and B. Product A sells at Rs 200 each and takes 1/2 hour to make. Product B sells at Rs 300 each and takes 1 hour to make. There is a permanent order for 14 units of product A and 16 units of product B. A working week consists of 40 hours of production and the weekly turn over must not be less than Rs 10000. If the profit on each of product A is Rs 20 and an product B is Rs 30, then how many of each should be produced so that the profit is maximum? Also find the maximum profit.
Answer:
Let x units of product A and y units of product B were manufactured.
Number of units cannot be negative.
Therefore, $x,y\ge 0$
According to question, the given information can be tabulated as:
Selling price(Rs)  Manufacturing time(hrs)  
Product A(x)  200  0.5 
Product B(y)  300  1 
Also, the availability of time is 40 hours and the revenue should be atleast Rs 10000.
Further, it is given that there is a permanent order for 14 units of product A and 16 units of product B.
Therefore, the constraints are
$200x+300y\ge 10000\phantom{\rule{0ex}{0ex}}0.5x+y\le 40\phantom{\rule{0ex}{0ex}}x\ge 14\phantom{\rule{0ex}{0ex}}y\ge 16$
If the profit on each of product A is Rs 20 and on product B is Rs 30.Therefore, profit gained on x units of product A and y units of product B is Rs 20x and Rs 30y respectively.
Total profit = Z = $20x+30y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = $20x+30y$
subject to
$2x+3y\ge 100\phantom{\rule{0ex}{0ex}}x+2y\le 80\phantom{\rule{0ex}{0ex}}x\ge 14\phantom{\rule{0ex}{0ex}}y\ge 16$
$x,y\ge 0$
First we will convert inequations into equations as follows:
2x + 3y = 100, x + 2y = 80, x = 14, y = 16, x = 0 and y = 0.
Region represented by 2x + 3y ≥ 100:
The line 2x + 3y = 100 meets the coordinate axes at A_{1}(50, 0) and ${B}_{1}\left(0,\frac{100}{3}\right)$ respectively. By joining these points we obtain the line
2x + 3y = 100 . Clearly (0,0) does not satisfies the 2x + 3y = 100. So, the region which does not contain the origin represents the solution set of the inequation 2x + 3y ≥ 100.
Region represented by x + 2y ≤ 80:
The line x + 2y = 80 meets the coordinate axes at C_{1}(80, 0) and D_{1}(0, 40) respectively. By joining these points we obtain the line
x + 2y = 80. Clearly (0,0) satisfies the inequation x + 2y ≤ 80. So,the region which contains the origin represents the solution set of the inequation x + 2y ≤ 80.
Region represented by x ≥ 14
x = 14 is the line passes through (14, 0) and is parallel to the Y axis.The region to the right of the line x = 14 will satisfy the inequation.
Region represented by y ≥ 16
y = 16 is the line passes through (0, 16) and is parallel to the X axis.The region above the line y = 16 will satisfy the inequation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 3y ≥ 100, x + 2y ≤ 80, x ≥ 14, y ≥ 16, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E_{1}(26, 16), F_{1}(48, 16), G_{1}(14, 33) and H_{1}(14, 24)
The values of Z at these corner points are as follows
Corner point  Z= 20x + 30y 
E_{1}  1000 
F_{1}  1440 
G_{1}  1270 
H_{1}  1000 
The maximum value of Z is Rs 1440 which is attained at F_{1}$\left(48,16\right)$.
Thus, the maximum profit is Rs 1440 obtained when 48 units of product A and 16 units of product B were manufactured.
Page No 30.55:
Question 36:
If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same. [CBSE 2007]
Answer:
Let us assume that the man travels x km when the speed is 25 km/hour and y km when the speed is 40 km/hour.
Thus, the total distance travelled is (x + y) km.
Now, it is given that the man has Rs 100 to spend on petrol.
Total cost of petrol = 2x + 5y ≤ 100
Now, time taken to travel x km = $\frac{x}{25}$ h
Time taken to travel y km = $\frac{y}{40}$ h
Now, it is given that the maximum time is 1 hour. So,
$\frac{x}{25}+\frac{y}{40}\le 1\phantom{\rule{0ex}{0ex}}\Rightarrow 8x+5y\le 200$
Thus, the given linear programming problem is
Maximise Z = x + y
subject to the constraints
2x + 5y ≤ 100
8x + 5y ≤ 200
x ≥ 0, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(25, 0), B$\left(\frac{50}{3},\frac{40}{3}\right)$ and C(0, 20).
The value of the objective function at these points are given in the following table.
Corner Points  Z = x + y 
(0, 0)  0 + 0 = 0 
(25, 0)  25 + 0 = 25 
$\left(\frac{50}{3},\frac{40}{3}\right)$  $\frac{50}{3}+\frac{40}{3}=30$ 
(0, 20)  0 + 20 = 20 
So, the maximum value of Z is 30 at $x=\frac{50}{3},y=\frac{40}{3}$.
Thus, the maximum distance that the man can travel in one hour is 30 km.
Hence, the distance travelled by the man at the speed of 25 km/hour is $\frac{50}{3}$ km, and the distance travelled by him at the speed of 40 km/hour is $\frac{40}{3}$km.
Page No 30.55:
Question 37:
An oil company has two depots, A and B, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps, D, E, F whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:
Figure
Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
Answer:
Let x and y litres of oil be supplied from A to the petrol pumps D and E. Then, (7000 − x − y) L will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L. Since, x L are transported from depot A, the remaining (4500 −x) L will be transported from petrol pump B.
Similarly, (3000 − y) L and [3500 − (7000 − x − y)] L i.e. (x + y − 3500) L will be transported from depot B to petrol pump E and F. respectively.
The given problem can be represented diagrammatically as follows.
Since, quantity of oil are nonnegative quantities.Therefore,
Cost of transporting 10 L of petrol = Re 1
Cost of transporting 1 L of petrol = $\mathrm{Rs}\frac{1}{10}$
Therefore, total transportation cost is given by,
The problem can be formulated as follows.
Minimize Z = 0.3x + 0.1y + 3950
subject to the constraints,
$x+y\le 7000\phantom{\rule{0ex}{0ex}}x\le 4500\phantom{\rule{0ex}{0ex}}y\le 3000\phantom{\rule{0ex}{0ex}}x+y\ge 3500\phantom{\rule{0ex}{0ex}}x,y\ge 0$
First we will convert inequations into equations as follows:
x + y = 7000, x = 4500, y = 3000, x + y = 3500, x = 0 and y = 0
Region represented by x + y ≤ 7000:
The line x + y = 7000 meets the coordinate axes at A_{1}(7000, 0) and ${B}_{1}\left(0,7000\right)$ respectively. By joining these points we obtain the line
x + y = 7000 . Clearly (0,0) satisfies the x + y = 7000 . So, the region which contains the origin represents the solution set of the inequation x + y ≤ 7000.
Region represented by x ≤ 4500:
The line x = 4500 is the line passes through C_{1}(4500, 0) and is parallel to Y axis. The region to the left of the line x = 4500 will satisfy the inequation
x ≤ 4500.
Region represented by y ≤ 3000:
The line y = 3000 is the line passes through D_{1}(0, 3000) and is parallel to X axis. The region below the the line y = 3000 will satisfy the inequation
y ≤ 3000.
Region represented by x + y ≥ 3500:
The line x + y = 7000 meets the coordinate axes at E_{1}(3500, 0) and ${F}_{1}\left(0,3500\right)$ respectively. By joining these points we obtain the line
x + y = 3500 . Clearly (0,0) satisfies the x + y = 3500. So, the region which contains the origin represents the solution set of the inequation x + y ≥ 3500.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 7000, x ≤ 4500, y ≤ 3000, x + y ≥ 3500, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E_{1}(3500, 0), C_{1}(4500, 0), I_{1}(4500, 2500), H_{1}(4000, 3000), and G_{1}(500, 3000).
The values of Z at these corner points are as follows.
Corner point  Z = 0.3x + 0.1y + 3950 
E_{1}(3500, 0)  5000 
C_{1}(4500, 0)  5300 
I_{1}(4500, 2500)  5550 
H_{1}(4000, 3000)  5450 
G_{1}(500, 3000)  4400 
The minimum value of Z is 4400 at G_{1}(500, 3000).
Thus, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively.
The minimum transportation cost is Rs 4400.
Page No 30.55:
Question 38:
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an LPP and solve it graphically.
Answer:
Let the firm manufacture x gold rings and y chains per day.
Number of gold rings and chains cannot be negative.
Therefore, x ≥0 and y ≥0.
It is given that the total number of gold rings and chains manufactured per day is at most 24.
∴ x + y ≤ 24
The gold ring takes 1 hour to make and chain takes 30 min, that is, 0.5 hour to make.
The maximum number of hours available per day is 16 hours.
∴ x + 0.5y ≤ 16
The profit on a ring is Rs 300 and on a chain is Rs 190. Therefore, profit made from x gold rings and y chains is Rs 300x and Rs 190y respectively.
∴ Total profit, Z = 300x + 190y
Thus, the mathematical formulation of the given problem is:
Maximise Z = 300x + 190y
subject to the constraints
x + y ≤ 24
x + 0.5y ≤ 16
x, y ≥0
First we will convert inequations into equations as follows:
x + y = 24, x + 0.5y = 16, x = 0 and y = 0
Region represented by x + y ≤ 24:
The line x + y = 24 meets the coordinate axes at A(24, 0) and $B\left(0,24\right)$ respectively. By joining these points we obtain the line x + y = 24. Clearly (0,0) satisfies the x + y = 24. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 24.
Region represented by x + 0.5y ≤ 16:
The line x + 0.5y = 16 meets the coordinate axes at C(16, 0) and $D\left(0,32\right)$ respectively. By joining these points we obtain the line x + 0.5y = 16. Clearly (0,0) satisfies the inequation x + 0.5y ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + 0.5y ≤ 16.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0 and y ≥ 0 are as follows.
The feasible region determined by the system constraints is as follows:
The corner points are O(0, 0), C(16, 0), E(8, 16) and B(0, 24).
The value of Z at these corner points are as follows:
Corner point 
Z = 300x + 190y 
O(0, 0)  0 
C(16, 0)  4800 
E(8, 16)  5440 
B(0, 24)  4560 
Thus, the maximum value of Z is 5440 at E(8, 16).
Thus, 8 gold rings and 16 chains should be manufactured per day to maximise the profits.
Page No 30.55:
Question 39:
A library has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weigh 1 kg and $1\frac{1}{2}$ kg each respectively. The shelf is 96 cm long and atmost can support a weight of 21 kg. How should the shelf be filled with the books of two types in order to include the greatest number of books? Make it as an LPP and solve it graphically.
Answer:
Let x books of first type and y books of second type were accommodated.
Number of books cannot be negative.
Therefore, $x,y\ge 0$
According to question, the given information can be tabulated as:
Thickness(cm)  Weight(kg)  
First type(x)  6  1 
Second type(y)  4  1.5 
Capacity of shelf  96  21 
Therefore, the constraints are
$6x+4y\le 96\phantom{\rule{0ex}{0ex}}x+1.5y\le 21$
Number of books = Z = $x+y$ which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
MaxImize Z = $x+y$
subject to
$6x+4y\le 96\phantom{\rule{0ex}{0ex}}x+1.5y\le 21$
First we will convert inequations into equations as follows:
6x + 4y = 96, x + 1.5y = 21, x = 0 and y = 0
Region represented by 6x + 4y ≤ 96:
The line 6x + 4y = 96 meets the coordinate axes at A(16, 0) and $B\left(0,24\right)$ respectively. By joining these points we obtain the line
6x + 4y = 96. Clearly (0,0) satisfies the 6x + 4y = 96. So, the region which contains the origin represents the solution set of the inequation
6x + 4y ≤ 96.
Region represented by x + 1.5y ≤ 21:
The line x + 1.5y = 21 meets the coordinate axes at C(21, 0) and $D\left(0,14\right)$ respectively. By joining these points we obtain the line
x + 1.5y = 21. Clearly (0,0) satisfies the inequation x + 1.5y ≤ 21. So,the region which contains the origin represents the solution set of the inequation x + 1.5y ≤ 21.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 6x + 4y ≤ 96, x + 1.5y ≤ 21, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), D(0, 14), E(12, 6), A(16, 0)
The values of Z at these corner points are as follows
Corner point  Z= x + y 
O  0 
D  14 
E  18 
A  16 
The maximum value of Z is 18 which is attained at E$\left(12,6\right)$.
Thus, maximum number of books that can be arranged on shelf is 18 where 12 books are of first type and 6 books are the other type.
Page No 30.55:
Question 40:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time. If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the number of tennis rackets and cricket bats that the factory must manufacture to earn the maximum profit. Make it as an LPP and solve it graphically.
Answer:
Let x number of tennis rackets and y number of cricket bats were sold.
Number of tennis rackets and cricket balls cannot be negative.
Therefore, $x\ge 0,y\ge 0$
It is given that a tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time.
Also, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.Therefore,
$1.5x+3y\le 42\phantom{\rule{0ex}{0ex}}3x+y\le 24$
If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively.Therefore, profit made on x tennis rackets and y cricket bats is Rs 20x and Rs 10y respectively.
Total profit = Z = 20x + 10y
Maximize Z = 20x + 10y
Subject to constraints:
$1.5x+3y\le 42\phantom{\rule{0ex}{0ex}}3x+y\le 24$
$x\ge 0,y\ge 0$
First we will convert inequations into equations as follows:
1.5x + 3y = 42, 3x + y = 24, x = 0 and y = 0
Region represented by 1.5x + 3y ≤ 42:
The line 1.5x + 3y = 42 meets the coordinate axes at A_{1}(28, 0) and ${B}_{1}\left(0,14\right)$ respectively. By joining these points we obtain the line 1.5x + 3y = 42. Clearly (0,0) satisfies the 1.5x + 3y = 42. So, the region which contains the origin represents the solution set of the inequation 1.5x + 3y ≤ 42.
Region represented by 3x + y ≤ 24:
The line 3x + y = 24 meets the coordinate axes at C_{1}(8, 0) and ${D}_{1}\left(0,24\right)$ respectively. By joining these points we obtain the line 3x + y = 24. Clearly (0,0) satisfies the inequation 3x + y ≤ 24. So,the region which contains the origin represents the solution set of the inequation 3x + y ≤ 24.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 1.5x + 3y ≤ 42, 3x + y ≤ 24, x ≥ 0 and y ≥ 0 are as follows.
In the above graph, the shaded region is the feasible region.
The corner points are O(0, 0), B_{1}(0, 14), E_{1}(4, 12), and C_{1}(8, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points 
Z = 20x +10y 

O(0, 0)  0  
B_{1}(0, 14) 
140 

E_{1}(4, 12) 
200 
Maximum 
C_{1}(8, 0) 
160 
Clearly, Z is maximum at x = 4 and y = 12 and the maximum value of Z at this point is 200.
Thus, maximum profit is of Rs 200 obtained when 4 tennis rackets and 12 cricket bats were sold.
Page No 30.55:
Question 41:
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Make an LPP and solve it graphically.
Answer:
Let x and y be the number of desktop model and portable model respectively.
Number of desktop model and portable model cannot be negative.
Therefore,
It is given that the monthly demand will not exist 250 units.
∴ $x+y\le 250$
Cost of desktop and portable model is Rs 25,000 and Rs 40,000 respectively. Therefore, cost of x desktop model and y portable model is Rs 25,000 and Rs 40,000 respectively and he does not want to invest more than Rs 70 lakhs.
$25000x+40000y\le 7000000\phantom{\rule{0ex}{0ex}}$
Profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Therefore, profit made by x desktop model and y portable model is Rs 4500x and Rs 5000y respectively.
Total profit = Z = 4500x + 5000y
Maximize Z = 4500x + 5000y
Subject to constraints:
$x+y\le 250$
$25000x+40000y\le 7000000\phantom{\rule{0ex}{0ex}}$
First we will convert inequations into equations as follows:
x + y = 250, 25000x + 40000y = 7000000, x = 0 and y = 0
Region represented by x + y ≤ 250:
The line x + y = 250 meets the coordinate axes at A(250, 0) and $B\left(0,250\right)$ respectively. By joining these points we obtain the line
x + y = 250. Clearly (0,0) satisfies the x + y = 250. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 250.
Region represented by 25000x + 40000y ≤ 7000000:
The line 25000x + 40000y = 7000000 meets the coordinate axes at C(280, 0) and $D\left(0,175\right)$ respectively. By joining these points we obtain the line 25000x + 40000y = 7000000. Clearly (0,0) satisfies the inequation 25000x + 40000y ≤ 7000000. So,the region which contains the origin represents the solution set of the inequation 25000x + 40000y ≤ 7000000.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points 
Z = 4500x + 5000y 

O(0, 0)  0  
D(0, 175) 
875000 

E(200, 50) 
1150000 
Maximum 
A(250, 0) 
1125000 
Clearly, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.
Thus, 200 desktop models and 50 portable units should be sold to maximize the profit.
Page No 30.55:
Question 42:
A cooperative society of farmers has 50 hectares of land to grow two crops X and Y. The profits from crops X and Y per hectare are estimated as ₹10,500 and ₹9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800 litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society? [CBSE 2013]
Answer:
Let the land allocated for crop X be x hectares and crop Y be y hectares.
Maximum area of the land available for two crops is 50 hectares.
∴ x + y ≤ 50
The liquid herbicide to be used for crops X and Y are at the rate of 20 litres and 10 litres per hectare, respectively. Maximum amount of herbicide to be used is 800 litres.
∴ 20x + 10 y ≤ 800
⇒ 2x + y ≤ 80
The profits from crops X and Y per hectare are ₹10,500 and ₹9,000, respectively.
Thus, total profit = ₹(10,500x + 9,000y)
Thus, the given linear programming problem is
Maximise Z = 10500x + 9000y
subject to the constraints
x + y ≤ 50
2x + y ≤ 80
x ≥ 0, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(40, 0), B(30, 20) and C(0, 50).
The value of the objective function at these points are given in the following table.
Corner Points  Z = 10500x + 9000y 
(0, 0)  10500 × 0 + 9000 × 0 = 0 
(40, 0)  10500 × 40 + 9000 × 0 = 420000 
(30, 20)  10500 × 30 + 9000 × 20 = 495000 
(0, 50)  10500 × 0 + 9000 × 50 = 450000 
So, the maximim value of Z is 495000 at x = 30 and y = 20.
Therefore, 30 hectares of land should be allocated for crop X and 20 hectares of land should be allocated for crop Y.
The maximum profit of the society is ₹4,95,000.
Page No 30.56:
Question 43:
A manufacturing company makes two models A and B of a product. Each piece of model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of ₹8000 on each piece of model A and ₹12000 on each piece of model B. How many pieces of model A and model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?
Answer:
Suppose x pieces of model A and y pieces of model B are manufactured per week.
Since each piece of model A requires 9 labour hours and each piece of model B requires 12 labour hours for fabricating, therefore, x pieces of model A and y pieces of model B require (9x + 12y) labour hours for fabricating. But, the maximum labour hours available for fabricating are 180.
∴ 9x + 12y ≤ 180
⇒ 3x + 4y ≤ 60
Similarly, each piece of model A requires 1 labour hour and each piece of model B requires 3 labour hours for finishing, therefore, x pieces of model A and y pieces of model B require (x + 3y) labour hours for finishing. But, the maximum labour hours available for finishing are 30.
∴ x + 3y ≤ 30
The profit from each piece of model A is ₹8,000 and from each piece of model B is ₹12,000. Therefore, the total profit from x pieces of model A and y pieces of model B is ₹(8,000x + 12,000y).
Thus, the given linear programming problem is
Maximise Z = 8000x + 12000y
subject to the constraints
3x + 4y ≤ 60
x + 3y ≤ 30
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(20, 0), B(12, 6) and C(0, 10).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 8000x + 12000y 
(0, 0)  8000 × 0 + 12000 × 0 = 0 
(20, 0)  8000 × 20 + 12000 × 0 = 160000 
(12, 6)  8000 × 12 + 12000 × 6 = 168000 → Maximum 
(0, 10)  8000 × 0 + 12000 × 10 = 120000 
The maximum value of Z is 168000 at x = 12, y = 6.
Hence, the manufacturing company should produce 12 pieces of model A and 6 pieces of model B to realise maximum profit. The maximum profit is ₹1,68,000.
Page No 30.56:
Question 44:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.
Answer:
Let x number of tennis rackets and y number of cricket bats were sold.
Number of tennis rackets and cricket balls cannot be negative.
Therefore, $x\ge 0,y\ge 0$
It is given that a tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time.
Also, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time. Therefore,
$1\xb75x+3y\le 42\phantom{\rule{0ex}{0ex}}3x+y\le 24$
If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively.Therefore, profit made on x tennis rackets and y cricket bats is Rs 20x and Rs 10y respectively.
Total profit = Z = 20x + 10y
The mathematical form of the given LPP is:
Maximize Z = 20x + 10y
Subject to constraints:
$1\xb75x+3y\le 42\phantom{\rule{0ex}{0ex}}3x+y\le 24$
$x\ge 0,y\ge 0$
First we will convert inequations into equations as follows:
1.5x + 3y = 42, 3x + y = 24, x = 0 and y = 0
Region represented by 1.5x + 3y ≤ 42:
The line 1.5x + 3y = 42 meets the coordinate axes at A(28, 0) and B(0, 14) respectively. By joining these points we obtain the line 1.5x + 3y = 42. Clearly, (0, 0) satisfies the 1.5x + 3y = 42. So, the region which contains the origin represents the solution set of the inequation 1.5x + 3y ≤ 42.
Region represented by 3x + y ≤ 24:
The line 3x + y = 24 meets the coordinate axes at C(8, 0) and D(0, 24) respectively. By joining these points we obtain the line 3x + y = 24. Clearly, (0, 0) satisfies the inequation 3x + y ≤ 24. So, the region which contains the origin represents the solution set of the inequation 3x + y ≤ 24.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 1.5x + 3y ≤ 42, 3x + y ≤ 24, x ≥ 0 and y ≥ 0 are as follows.
In the above graph, the shaded region is the feasible region.
The corner points are O(0, 0), B(0, 14), E(4, 12), and C(8, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points 
Z = 20x +10y 

O(0, 0)  0  
B(0, 14) 
140 

E(4, 12) 
200 
Maximum 
C(8, 0) 
160 
Clearly, Z is maximum at x = 4 and y = 12 and the maximum value of Z at this point is 200.
(i) 4 tennis rackets and 12 cricket bats must be made if the factory is to work at full capacity.
(ii) The maximum profit of the factory when it works at full capacity is Rs 200.
Page No 30.56:
Question 45:
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and his profit on the desktop model is Rs 4500 and on the portable model is Rs 5000.
Answer:
Let x and y be the number of desktop model and portable model respectively.
Number of desktop model and portable model cannot be negative.
Therefore,
It is given that the monthly demand will not exist 250 units.
∴ $x+y\le 250$
Cost of desktop and portable model is Rs 25,000 and Rs 40,000 respectively. Therefore, cost of x desktop model and y portable model is Rs 25,000 and Rs 40,000 respectively and he does not want to invest more than Rs 70 lakhs.
$25000x+40000y\le 7000000\phantom{\rule{0ex}{0ex}}$
Profit on the desktop model is Rs 4500 and on the portable model is Rs 5000. Therefore, profit made by x desktop model and y portable model is Rs 4500x and Rs 5000y respectively.
Total profit = Z = 4500x + 5000y
Maximize Z = 4500x + 5000y
Subject to constraints:
$x+y\le 250$
$25000x+40000y\le 7000000\phantom{\rule{0ex}{0ex}}$
First we will convert inequations into equations as follows:
x + y = 250, 25000x + 40000y = 7000000, x = 0 and y = 0
Region represented by x + y ≤ 250:
The line x + y = 250 meets the coordinate axes at A(250, 0) and $B\left(0,250\right)$ respectively. By joining these points we obtain the line
x + y = 250. Clearly (0,0) satisfies the x + y = 250. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 250.
Region represented by 25000x + 40000y ≤ 7000000:
The line 25000x + 40000y = 7000000 meets the coordinate axes at C(280, 0) and $D\left(0,175\right)$ respectively. By joining these points we obtain the line 25000x + 40000y = 7000000. Clearly (0,0) satisfies the inequation 25000x + 40000y ≤ 7000000. So,the region which contains the origin represents the solution set of the inequation 25000x + 40000y ≤ 7000000.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows.
The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points 
Z = 4500x + 5000y 

O(0, 0)  0  
D(0, 175) 
875000 

E(200, 50) 
1150000 
Maximum 
A(250, 0) 
1125000 
Clearly, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.
Thus, 200 desktop models and 50 portable units should be sold to maximize the profit.
Page No 30.56:
Question 47:
There are two types of fertilizers F_{1 }and F_{2}. F_{1 }consists of 10% nitrogen and 6% phosphoric acid and F_{2 }consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F_{1 }costs ₹6/kg and F_{2 }costs ₹5/kg, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost?
Answer:
Suppose x kg of fertilizer F_{1 }and and y kg of fertilizer F_{2} is used to meet the nutrient requirements.
F_{1 }consists of 10% nitrogen and F_{2 }consists of 5% nitrogen. But, the farmer needs atleast 14 kg of nitorgen for the crops.
∴ 10% of x kg + 5% of y kg ≥ 14 kg
$\Rightarrow \frac{x}{10}+\frac{y}{20}\ge 14\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+y\ge 280$
Similarly, F_{1 }consists of 6% phosphoric acid and F_{2 }consists of 10% phosphoric acid. But, the farmer needs atleast 14 kg of phosphoric acid for the crops.
∴ 6% of x kg + 10% of y kg ≥ 14 kg
$\Rightarrow \frac{6x}{100}+\frac{10y}{100}\ge 14\phantom{\rule{0ex}{0ex}}\Rightarrow 3x+5y\ge 700$
The cost of fertilizer F_{1 }is ₹6/kg and fertilizer F_{2 }is ₹5/kg, therefore, total cost of x kg of fertilizer F_{1 }and and y kg of fertilizer F_{2 }is ₹(6x + 5y).
Thus, the given linear programming problem is
Minimise Z = 6x + 5y
subject to the constraints
2x + y ≥ 280
3x + 5y ≥ 700
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are $\mathrm{A}\left(\frac{700}{3},0\right)$, B(100, 80) and C(0, 280).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 6x + 5y 
$\left(\frac{700}{3},0\right)$  $6\times \frac{700}{3}+5\times 0=1400$ 
(100, 80)  6 × 100 + 5 × 80 = 1000 → Minimum 
(0, 280)  6 × 0 + 5 × 280 = 1400 
The smallest value of Z is 1000 which is obtained at x = 100, y = 80.
It can be seen that the open halfplane represented by 6x + 5y < 1000 has no common points with the feasible region.
So, the minimum value of Z is 1000.
Hence, 100 kg of fertilizer F_{1 }and and 80 kg of fertilizer F_{2} should be used so that the nutrient requirements are met at minimum cost. The minimum cost is ₹1,000.
Page No 30.56:
Question 48:
A manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit each of M and N on the three machines are given in the following table:
Items  Number of hours required on machines  
I  II  III  
M  1  2  1 
N  2  1  1.25 
She makes a profit of ₹600 and ₹400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
Answer:
Suppose x units of item M and y units of item N are produced to maximise the profit.
Since each unit of item M require 1 hours on machine I and each unit of item N require 2 hours on machine I, therefore, the total hours required for producing x units of item M and y units of item N on machine I are (2x + y). But, machines I is capable of being operated for at most 12 hours.
∴ 2x + y ≤ 12
Similarly, each unit of item M require 2 hours on machine II and each unit of item N require 1 hour on machine II, therefore, the total hours required for producing x units of item M and y units of item N on machine II are (x + 2y). But, machines II is capable of being operated for at most 12 hours.
∴ x + 2y ≤ 12
Also, each unit of item M require 1 hour on machine III and each unit of item N require 1.25 hour on machine III, therefore, the total hours required for producing x units of item M and y units of item N on machine III are (x + 1.25y). But, machines III must be operated for atleast 5 hours.
∴ x + 1.25y ≥ 5
The profit from each unit of item M is ₹600 and each unit of item N is ₹400. Therefore, the total profit from x units of item M and y units of item N is ₹(600x + 400y).
Thus, the given linear programming problem is
Maximise Z = 600x + 400y
subject to the constraints
2x + y ≤ 12
x + 2y ≤ 12
x + 1.25y ≥ 5
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(5, 0), B(6, 0), C(4, 4), D(0, 6) and E(0, 4).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 600x + 400y 
(5, 0)  600 × 5 + 400 × 0 = 3000 
(6, 0)  600 × 6 + 400 × 0 = 3600 
(4, 4)  600 × 4 + 400 × 4 = 4000 → Maximum 
(0, 6)  600 × 0 + 400 × 6 = 2400 
(0, 4)  600 × 0 + 400 × 4 = 1600 
The maximum value of Z is 4000 at x = 4, y = 4.
Hence, 4 units of item M and 4 units of item N should be produced to maximise the profit. The maximum profit of the manufacturer is ₹4,000.
Page No 30.57:
Question 49:
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:
From \ To  Cost (in ₹)  
A  B  C  
P  160  100  150 
Q  100  120  100 
How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?
Answer:
Here, demand of the commodity (5 + 5 + 4 = 14 units) is equal the supply of the commodity (8 + 6 = 14 units). So, no commodity would be left at the two factories.
Let x units and y units of the commodity be transported from the factory P to the depots at A and B, respectively.
Then, (8 − x − y) units of the commodity will be transported from the factory P to the depot C.
Now, the weekly requirement of depot A is 5 units of the commodity. Now, x units of the commodity are transported from factory P, so the remaining (5 − x) units of the commodity are transported from the factory Q to the depot A.
The weekly requirement of depot B is 5 units of the commodity. Now, y units of the commodity are transported from factory P, so the remaining (5 − y) units of the commodity are transported from the factory Q to the depot B.
Similarly, 6 − (5 − x) − (5 − y) = (x + y − 4) units of the commodity will be transported from the factory Q to the depot C.
Since the number of units of commodity transported are from the factories to the depots are nonnegative, therefore,
x ≥ 0, y ≥ 0, 8 − x − y ≥ 0, 5 − x ≥ 0, 5 − y ≥ 0, x + y − 4 ≥ 0
Or x ≥ 0, y ≥ 0, x + y ≤ 8, x ≤ 5, y ≤ 5, x + y ≥ 4
Total transportation cost = 160x + 100y + 150(8 − x − y) + 100(5 − x) + 120(5 − y) + 100(x + y − 4) = 10x − 70y + 1900
Thus, the given linear programming problem is
Minimise Z = 10x − 70y + 1900
subject to the constraints
x + y ≤ 8
x ≤ 5
y ≤ 5
x + y ≥ 4
x ≥ 0, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(4, 0), B(5, 0), C(5, 3), D(3, 5), E(0, 5) and F(0, 4).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 10x − 70y + 1900 
(4, 0)  10 × 4 − 70 × 0 + 1900 = 1940 
(5, 0)  10 × 5 − 70 × 0 + 1900 = 1950 
(5, 3)  10 × 5 − 70 × 3 + 1900 = 1740 
(3, 5)  10 × 3 − 70 × 5 + 1900 = 1580 
(0, 5)  10 × 0 − 70 × 5 + 1900 = 1550 → Minimum 
(0, 4)  10 × 0 − 70 × 4 + 1900 = 1620 
The minimum value of Z is 1550 at x = 0, y = 5.
Hence, for minimum transportation cost, factory P should supply 0, 5, 3 units of commodity to depots A, B, C respectively and factory Q should supply 5, 0, 1 units of commodity to depots A, B, C respectively.
The minimum transportation cost is ₹1,550.
Page No 30.57:
Question 50:
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
Types of Toys  Machines  
I  II  III  
A  12  18  6 
B  6  0  9 
Answer:
Suppose the manufacturer makes x toys of type A and y toys of type B.
Since each toy of type A require 12 minutes on machine I and each toy of type B require 6 minutes on machine I, therefore, x toys of type A and y toys of type B require (12x + 6y) minutes on machine I. But, machines I is available for at most 6 hours.
∴ 12x + 6y ≤ 360
⇒ 2x + y ≤ 60
Similarly, each toy of type A require 18 minutes on machine II and each toy of type B require 0 minutes on machine II, therefore, x toys of type A and y toys of type B require (18x + 0y) minutes on machine II. But, machines II is available for at most 6 hours.
∴ 18x + 0y ≤ 360
⇒ x ≤ 20
Also, each toy of type A require 6 minutes on machine III and each toy of type B require 9 minutes on machine III, therefore, x toys of type A and y toys of type B require (6x + 9y) minutes on machine III. But, machines III is available for at most 6 hours.
∴ 6x + 9y ≤ 360
⇒ 2x + 3y ≤ 120
The profit on each toy of type A is ₹7.50 and each toy of type B is ₹5. Therefore, the total profit from x toys of type A and y toys of type B is ₹(7.50x + 5y).
Thus, the given linear programming problem is
Maximise Z = 7.5x + 5y
subject to the constraints
2x + y ≤ 60
x ≤ 20
2x + 3y ≤ 120
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(20, 0), B(20, 20), C(15, 30) and D(0, 40).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 7.5x + 5y 
(0, 0)  7.5 × 0 + 5 × 0 = 0 
(20, 0)  7.5 × 20 + 5 × 0 = 150 
(20, 20)  7.5 × 20 + 5 × 20 = 250 
(15, 30)  7.5 × 15 + 5 × 30 = 262.5 → Maximum 
(0, 40)  7.5 × 0 + 5 × 40 = 200 
The maximum value of Z is 262.5 at x = 15, y = 30.
Hence, 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit. The maximum profit is ₹262.50.
Page No 30.57:
Question 51:
An aeroplane can carry a maximum of 200 passengers. A profit of ₹1000 is made on each executive class ticket and a profit of ₹600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit of the airline. What is the maximum profit?
Answer:
Suppose x tickets of executive class and y tickets of economy class are sold by the airline.
The profit on each executive class ticket is ₹1000 and on each economy class ticket is ₹600. Therefore, the total profit from x executive class tickets and y economy class ticket is ₹(1000x + 600y).
Now, the aeroplane can carry a maximum of 200 passengers.
∴ x + y ≤ 200
The airline reserves atleast 20 seats for executive class.
∴ x ≥ 20
Also, atleast 4 times as many passengers prefer to travel by economy class than by the executive class.
∴ y ≥ 4x
Thus, the given linear programming problem is
Maximise Z = 1000x + 600y
subject to the constraints
x + y ≤ 200
x ≥ 20
y ≥ 4x
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(20, 80), B(40, 160) and C(20, 180).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 1000x + 600y 
(20, 80)  1000 × 20 + 600 × 80 = 68000 
(40, 160)  1000 × 40 + 600 × 160 = 136000 → Maximum 
(20, 180)  1000 × 20 + 600 × 180 = 128000 
The maximum value of Z is 136000 at x = 40, y = 160.
Hence, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit. The maximum profit of the airline is ₹1,36,000.
Page No 30.57:
Question 52:
A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at ₹100 and ₹120 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?
Answer:
Let x units of A and y units of B be produced by the manufacturer.
The price of one unit of A is ₹100 and the price of one unit of B is ₹120. Therefore, the total price of x units of A and y units of B or the total revenue is ₹(100x + 120y).
One unit of A requires 2 workers and one unit of B requires 3 workers. Therefore, x units of A and y units of B requires (2x + 3y) workers. But, the manufacturer has 30 workers.
∴ 2x + 3y ≤ 30
Similarly, one unit of A requires 3 units of capital and one unit of B requires 1 unit of capital. Therefore, x units of A and y units of B requires (3x + y) units of capital. But, the manufacturer has 17 units of capital.
∴ 3x + y ≤ 17
Thus, the given linear programming problem is
Maximise Z = 100x + 120y
subject to the constraints
2x + 3y ≤ 30
3x + y ≤ 17
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(0, 10), B$\left(\frac{17}{3},0\right)$ and C(3, 8).
The value of the objective function at these points are given in the following table.
Corner Point  Z = 100x + 120y 
(0, 0)  100 × 0 + 120 × 0 = 0 
(0, 10)  100 × 0 + 120 × 10 = 1200 
$\left(\frac{17}{3},0\right)$  100 × $\frac{17}{3}$ + 120 × 0 = $\frac{1700}{3}$ 
(3, 8)  100 × 3 + 120 × 8 = 1260 → Maximum 
The maximum value of Z is 1260 at x = 3, y = 8.
Hence, the maximum total revenue is ₹1,260 when 3 units of A and 8 units of B are produced.
Yes, because the efficiency of a worker does not depend on whether the worker is a male or a female.
Page No 30.57:
Question 53:
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at ₹7 profit and that of B at a profit of ₹4. Find the production level per day for maximum profit graphically.
Answer:
Let x units of product A and y units of product B be manufactured by the manufacturer per day.
It is given that one unit of product A requires 3 hours of processing time on first machine, while one unit of product B requires 2 hours of processing time on first machine. It is also given that first machine is available for 12 hours per day.
∴ 3x + 2y ≤ 12
Also, one unit of product A requires 3 hours of processing time on second machine, while one unit of product B requires 1 hour of processing time on second machine. It is also given that second machine is available for 9 hours per day.
∴ 3x + y ≤ 9
The profits on one unit each of product A and product B is ₹ 7 and ₹ 4, respectively.
So, the objective function is given by Z = ₹ (7x + 4y).
Therefore, the mathematical formulation of the given linear programming problem can be stated as:
Maximize Z = 7x + 4y
Subject to the constraints
3x + 2y ≤ 12 .....(1)
3x + y ≤ 9 .....(2)
x ≥ 0, y ≥ 0 .....(3)
The feasible region determined by constraints (1) to (3) is graphically represented as:
Here, it is seen that OABCO is the feasible region and it is bounded. The values of Z at the corner points of the feasible region are represented in tabular form as:
Corner Point  Z = 7x + 4y 
O(0, 0)  Z = 7 × 0 + 4 × 0 = 0 
A(3, 0)  Z = 7 × 3 + 4 × 0 = 21 
B(2, 3)  Z = 7 × 2 + 4 × 3 = 26 
C(0, 6)  Z = 7 × 0 + 4 × 6 = 24 
The maximum value of Z is 26, which is obtained at x = 2 and y = 3.
Thus, 2 units of product A and 3 units of product B should be manufactured by the manufacturer per day in order to maximize the profit.
Also, the maximum daily profit of the manufacturer is ₹ 26.
Page No 30.57:
Question 54:
There are two types of fertilisers 'A' and 'B' . 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs ₹10 per kg and 'B' cost ₹8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost
Answer:
The given information can tabulated as follows:
Fertilizer  Nitrogen  Phosphoric Acid  Cost/kg (in ₹) 
A  12%  5%  10 
B  4%  5%  8 
Let the requirement of fertilizer A by the farmer be x kg and that of B be y kg.
It is given that farmer requires atleast 12 kg of nitrogen and 12 kg of phosphoric acid for his crops.
The inequations thus formed based on the given information are as follows:
$\frac{12}{100}x+\frac{4}{100}y\ge 12\phantom{\rule{0ex}{0ex}}\Rightarrow 12x+4y\ge 1200\phantom{\rule{0ex}{0ex}}\Rightarrow 3x+y\ge 300.....\left(1\right)$
Also,
$\frac{5x}{100}+\frac{5y}{100}\ge 12\phantom{\rule{0ex}{0ex}}\Rightarrow 5x+5y\ge 1200\phantom{\rule{0ex}{0ex}}\Rightarrow x+y\ge 240.....\left(2\right)$
Total cost of the fertilizer Z = ₹ (10x + 8y)
Therefore, the mathematical formulation of the given linear programming problem can be stated as:
Minimize Z = 10x + 8y
Subject to the constraints
3x + y ≥ 300 .....(1)
x + y ≥ 240 .....(2)
x ≥ 0, y ≥ 0 .....(3)
The feasible region determined by constraints (1) to (3) is graphically represented as:
Here, it is seen that the feasible region is unbounded. The values of Z at the corner points of the feasible region are represented in tabular form as:
Corner Point  Z = 10x + 8y 
A(0, 300)  Z = 10 × 0 + 8 × 300 = 2400 
B(30, 210)  Z = 10 × 30 + 8 × 210 = 1980 
C(240, 0)  Z = 10 × 240 + 8 × 0 = 2400 
The open half plane determined by 10x + 8y < 1980 has no point in common with the feasible region. So, the minimum value of Z is 1980.
The minimum value of Z is 1980, which is obtained at x = 30 and y = 210.
Thus, the minimum requirement of fertilizer of type A will be 30 kg and that of type B will be 210 kg.
Also, the total minimum cost of the fertilisers is ₹ 1980.
Page No 30.57:
Question 55:
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?
It is being given that at least one of each must be produced.
Answer:
Let the number of necklaces manufacture be x,
and the number of bracelets manufacture be y.
since the total number of items are at most 24.
$x+y\le 24$ ...(1)
Bracelets takes 1 hour to manufacture and necklaces takes half an hour to manufacture.
x item takes x hour to manufacture and y items take y/2 hour to manufacture.
and maximum time available is 16 hours.
therefore
$\frac{x}{2}+y\le 16$ ...(2)
the profit on one necklace is Rs. 100 and the profit on one bracelet is Rs.300
Let the profit be Z. Now we wish to maximize the profit. So,
Max Z = 100x + 300y ...(3)
So, $x+y\le 24$,
$\frac{x}{2}+y\le 16$
Max Z = 100x + 300y
is the required L.P.P.
Page No 30.65:
Question 1:
Tow godowns, A and B, have grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
Transportation cost per quintal(in Rs.)  
From>  A  B 
To  
D  6.00  4.00 
E  3.00  2.00 
F  2.50  3.00 
How should the supplies be transported in order that the transportation cost is minimum?
Answer:
Let godown A supply x quintals and y quintals of grain to the shops D and E respectively.
Then, (100 − x − y) will be supplied to shop F.
The requirement at shop D is 60 quintals since, x quintals are transported from godown A.
Therefore, the remaining (60 − x) quintals will be transported from godown B.
Similarly, (50 − y) quintals and 40 − (100 − x − y) i.e. (x + y − 60) quintals will be transported from godown B to shop E and F respectively.
The given problem can be represented diagrammatically as follows.
Quantity of the grain cannot be negative.Therefore,
Total transportation cost Z is given by,
$\mathrm{Z}=6x+3y+2.5\left(100xy\right)+4\left(60x\right)+2\left(50y\right)+3\left(x+y60\right)\phantom{\rule{0ex}{0ex}}=6x+3y+2502.5x2.5y+2404x+1002y+3x+3y180\phantom{\rule{0ex}{0ex}}=2.5x+1.5y+410$
The given problem can be formulated as:
Minimize Z = 2.5x + 1.5y + 410
subject to the constraints,
$x+y\le 100\phantom{\rule{0ex}{0ex}}x\le 60\phantom{\rule{0ex}{0ex}}y\le 50\phantom{\rule{0ex}{0ex}}x+y\ge 60\phantom{\rule{0ex}{0ex}}x,y\ge 0$
First we will convert inequations into equations as follows:
x + y = 100, x = 60, y = 50, x + y =60, x = 0 and y = 0
Region represented by x + y ≤ 100:
The line x + y = 100 meets the coordinate axes at A_{1}(100, 0) and B_{1}(0, 100) respectively. By joining these points we obtain the line x + y = 100. Clearly (0,0) satisfies the x + y = 100. So, the region which contains the origin represents the solution set of the inequation x + y ≤ 100.
Region represented by x ≤ 60:
x = 60 is the line that passes (60, 0) and is parallel to the Y axis.The region to the left of the line x = 60 will satisfy the inequation x ≤ 60.
Region represented by y ≤ 50:
y = 50 is the line that passes (0, 50) and is parallel to the X axis.The region below the line y = 50 will satisfy the inequation y ≤ 50.
Region represented by x + y ≥ 60:
The line x + y = 60 meets the coordinate axes at C_{1}(60, 0) and ${D}_{1}\left(0,60\right)$ respectively. By joining these points we obtain the line x + y = 60. Clearly (0,0) does not satisfies the inequation x + y ≥ 60. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 60.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 100, x ≤ 60, y ≤ 50, x + y ≥ 60, x ≥ 0 and y ≥ 0 are as follows.
The corner points are C_{1}(60, 0), G_{1}(60, 40), F_{1}(50, 50), and E_{1}(10, 50).
The values of Z at these corner points are as follows.
Corner point  Z = 2.5x + 1.5y + 410 
C_{1}(60, 0)  560 
G_{1}(60, 40)  620 
F_{1}(50, 50)  610 
E_{1}(10, 50)  510 
The minimum value of Z is 510 at E_{1}(10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is Rs 510.
Page No 30.65:
Question 2:
A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B, are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:
Transportation Cost per packet(in Rs.)  
From>  A  B 
To  
P  5  4 
Q  4  2 
R  3  5 
How many packets from each factory be transported to each agency so that the cost of transportation is minimum? Also find the minimum cost?
Answer:
Let x and y packets be transported from factory A to the agencies P and Q respectively. Then, [60 − (x + y)] packets be transported to the agency R.
The requirement at agency P is 40 packets. Since, x packets are transported from factory A,
Therefore, the remaining (40 − x) packets are transported from factory B.
Similarly, (40 − y) packets are transported by B to Q and 50− [60 − (x + y)] i.e. (x + y − 10) packets will be transported from factory B to agency R respectively.
Number of packets cannot be negative.Therefore,
$x\ge 0,y\ge 0\mathrm{and}60xy\ge 0\phantom{\rule{0ex}{0ex}}\Rightarrow x\ge 0,y\ge 0\mathrm{and}x+y\le 60\phantom{\rule{0ex}{0ex}}40x\ge 0,40y\ge 0\mathrm{and}x+y10\ge 0\phantom{\rule{0ex}{0ex}}\Rightarrow x\mathit{\le}40,y\le 40\mathrm{and}x+y\ge 10$
Total transportation cost Z is given by,
$\mathrm{Z}=5x+4y+3\left[60\left(x+y\right)\right]+4\left(40x\right)+2\left(40y\right)+5\left(x+y10\right)\phantom{\rule{0ex}{0ex}}=3x+4y+10$
Minimize Z = $5x+4y+3\left(60xy\right)+4\left(40x\right)+2\left(40y\right)+5\left(x+y10\right)$
=$3x+4y+370$
subject to
$x+y\le 60\phantom{\rule{0ex}{0ex}}x\le 40\phantom{\rule{0ex}{0ex}}y\le 40\phantom{\rule{0ex}{0ex}}x+y\ge 10\phantom{\rule{0ex}{0ex}}x,y\ge 0$
First we will convert inequations into equations as follows:
x + y = 60, x = 40, y = 40, x + y = 10, x = 0 and y = 0
Region represented by x + y ≤ 60:
The line x + y = 60 meets the coordinate axes at A_{1}(60, 0) and B_{1}(0, 60) respectively. By joining these points we obtain the line
x + y = 60. Clearly (0,0) satisfies the x + y = 60. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 60.
Region represented by x ≤ 40:
x = 40 is the line that passes C_{1}(40, 0) and is parallel to the Y axis.The region to the left of the line x = 40 will satisfy the inequation x ≤ 40.
Region represented by y ≤ 40:
y = 40 is the line that passes D_{1}(0, 40) and is parallel to the X axis.The region below the line y = 40 will satisfy the inequation y ≤ 40.
Region represented by x + y ≥ 10:
The line x + y = 10 meets the coordinate axes at E_{1}(10, 0) and ${F}_{1}\left(0,10\right)$ respectively. By joining these points we obtain the line x + y = 10. Clearly (0,0) does not satisfies the inequation x + y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 10.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + y ≤ 60, x ≤ 40, y ≤ 40, x + y ≥ 10, x ≥ 0 and y ≥ 0 are as follows.
The corner points are D_{1}(0, 40), H_{1}(20, 40), G_{1}(40, 20), C_{1}(40, 0), E_{1}(10, 0) and F_{1}(0, 10).
The values of Z at these corner points are as follows
Corner point  Z= 3x + 4y + 370 
D_{1}  530 
H_{1}  590 
G_{1}  570 
C_{1}  490 
E_{1}  400 
F_{1}  410 
The minimum value of Z is 400 which is at E_{1}(10, 0).
Thus, the minimum cost is Rs 400.
Hence,
From A: 10 packets, 0 packets and 50 packets to P, Q and R respectively
From B: 30 packets, 40 packets and 0 packets to P, Q and R respectively
Page No 30.67:
Question 1:
The solution set of the inequation 2x + y > 5 is
(a) half plane that contains the origin
(b) open half plane not containing the origin
(c) whole xyplane except the points lying on the line 2x + y = 5
(d) none of these
Answer:
(b) open half plane not containing the origin
On putting x = 0, y = 0 in the given inequality, we get
0 > 5, which is absurd.
Therefore, the solution set of the given inequality does not include the origin.
Thus, the solution set of the given inequality consists of the open half plane not containing the origin.
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Question 2:
Objective function of a LPP is
(a) a constraint
(b) a function to be optimized
(c) a relation between the variables
(d) none of these
Answer:
(b) a function to be optimized
The objective function of a linear programming problem is either to be maximized or minimized i.e. objective function is to be optimized
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Question 3:
Which of the following sets are convex?
(a) {(x, y) : x^{2} + y^{2}^{ }≥ 1}
(b) {(x, y) : y^{2} ≥ x}
(c) {(x, y) : 3x^{2} + 4y^{2} ≥ 5}
(d) {(x, y) : y ≥ 2, y ≤ 4}
Answer:
(d) {(x, y) : y ≥ 2, y ≤ 4}
{(x, y) : y ≥ 2, y ≤ 4} is the region between two parallel lines, so any line segment joining any two points in it lies in it. Hence, it is a convex set.
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Question 4:
Let X_{1} and X_{2} are optimal solutions of a LPP, then
(a) X = λ X_{1} + (1 − λ) X_{2}, λ ∈ R is also an optimal solution
(b) X = λ X_{1} + (1 − λ) X_{2}, 0 ≤ λ ≤ 1 gives an optimal solution
(c) X = λ X_{1} + (1 + λ) X_{2}, 0 ≤ λ ≤ 1 gives an optimal solution
(d) X = λ X_{1} + (1 + λ) X_{2}, λ ∈ R gives an optimal solution
Answer:
(b) X = λX_{1} + (1 − λ)X_{2}, 0 ≤ λ ≤ 1 gives an optimal solution
A set A is convex if, for any two points, x_{1}, x_{2} ∈ A, and $\mathrm{\lambda}\in \left[0,1\right]$ imply that
$\lambda {x}_{1}+\left(1\lambda \right){x}_{2}\in A$.
Since,here X_{1}_{ }and X_{2} are optimal solutions
Therefore, their convex combination will also be an optimal solution
Thus, X = λX_{1} + (1 − λ)X_{2}, 0 ≤ λ ≤ 1 gives an optimal solution.
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Question 5:
The maximum value of Z = 4x + 2y subjected to the constraints 2x + 3y ≤ 18, x + y ≥ 10; x, y ≥ 0 is
(a) 36
(b) 40
(c) 20
(d) none of these
Answer:
(d) none of these
We need to maximize the function Z = 4x + 2y
Converting the given inequations into equations, we obtain
$2x+3y=18,x+y=10,x=0\mathrm{and}y=0$
Region represented by 2x + 3y ≤ 18:
The line 2x + 3y = 18 meets the coordinate axes at A(9, 0) and B(0, 6) respectively. By joining these points we obtain the line 2x + 3y = 18.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 18. So,the region in xy plane which contain the origin represents the solution set of the inequation 2x + 3y ≤ 18.
Region represented by x + y ≥ 10:
The line x + y = 10 meets the coordinate axes at $C\left(10,0\right)$ and D(0, 10) respectively. By joining these points we obtain the line x + y =10.
Clearly (0,0) does not satisfies the inequation x + y ≥ 10. So,the region which does not contain the origin represents the solution set of the inequation x + y ≥ 10.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
We observe that feasible region of the given LPP does not exist.
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Question 6:
The optimal value of the objective function is attained at the points
(a) given by intersection of inequations with the axes only
(b) given by intersection of inequations with xaxis only
(c) given by corner points of the feasible region
(d) none of these
Answer:
(c) given by corner points of the feasible region
It is known that the optimal value of the objective function is attained at any of the corner point.
Thus, the optimal value of the objective function is attained at the points given by corner points of the feasible region.
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Question 7:
The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80; x, y ≥ 0 is
(a) 320
(b) 300
(c) 230
(d) none of these
Answer:
(d) none of these
We need to maximize the function Z = 4x + 3y
Converting the given inequations into equations, we obtain
$3x+2y=160,5x+2y=200,x+2y=80,x=0\mathrm{and}y=0$
Region represented by 3x + 2y ≥ 160:
The line 3x + 2y = 160 meets the coordinate axes at $A\left(\frac{160}{3},0\right)$ and B(0, 80) respectively. By joining these points we obtain the line
3x + 2y = 160.Clearly (0,0) does not satisfies the inequation 3x + 2y ≥ 160. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 3x + 2y ≥ 160.
Region represented by 5x +2y ≥ 200:
The line 5x +2y = 200 meets the coordinate axes at C(40,0) and D(0, 100) respectively. By joining these points we obtain the line
5x +2y = 200.Clearly (0,0) does not satisfies the inequation 5x +2y ≥ 200. So,the region which does not contain the origin represents the solution set of the inequation 5x +2y ≥ 200.
Region represented by x +2y ≥ 80:
The line x +2y = 80 meets the coordinate axes at E(80,0) and F(0, 40) respectively. By joining these points we obtain the line
x +2y = 80.Clearly (0,0) does not satisfies the inequation x +2y ≥ 80. So,the region which does not contain the origin represents the solution set of the inequation x +2y ≥ 80.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 3x + 2y ≥ 160, 5x +2y ≥ 200, x +2y ≥ 80, x ≥ 0, and y ≥ 0 are as follows.
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Question 8:
Consider a LPP given by
Minimum Z = 6x + 10y
Subjected to x ≥ 6; y ≥ 2; 2x + y ≥ 10; x, y ≥ 0
Redundant constraints in this LPP are
(a) x ≥ 0, y ≥ 0
(b) x ≥ 6, 2x + y ≥ 10
(c) 2x + y ≥ 10
(d) none of these
Answer:
(c) $2x+y\ge 10$
We need to minimize the function Z = 6x + 10y
Converting the given inequations into equations, we obtain
$x=6,y=2,2x+y=10,x=0,y=0$
These lines are drawn using a suitable scale
Scale
On X axis
1 Big division = 1 unit
On Y axis
1 Big division = 1 unit
The shaded region represents the feasible region of the given LPP.
We observe that the feasible region is due to the constraints $x\ge 6,y\ge 2$
So, the redundant constraint is $2x+y\ge 10$
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Question 9:
The objective function Z = 4x + 3y can be maximised subjected to the constraints 3x + 4y ≤ 24, 8x + 6y ≤ 48, x ≤ 5, y ≤ 6; x, y ≥ 0
(a) at only one point
(b) at two points only
(c) at an infinite number of points
(d) none of these
Answer:
(c) at an infinite number of points
We need to maximize Z = 4x + 3y
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5 , y = 6, x = 0 and y = 0.
The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0,6). Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation 3x + 4y ≤ 24.So, the region in xyplane that contains the origin represents the solution set of the given equation.
The line 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0,8). Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation 8x + 6y ≤ 48. So, the region in xyplane that contains the origin represents the solution set of the given equation.
x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
GRAPH
The corner points of the feasible region are O(0, 0), $G\left(5,0\right)$, $F\left(5,\frac{4}{3}\right)$, $E\left(\frac{24}{7},\frac{24}{7}\right)$ and $B\left(0,6\right)$.
The values of Z at these corner points are as follows.
Corner point  Z = 4x + 3y 
O(0, 0)  4× 0 + 3 × 0 = 0 
$G\left(5,0\right)$  4 × 5 + 3 × 0 = 20 
$F\left(5,\frac{4}{3}\right)$  4 × 5 + 3 ×$\frac{4}{3}$ = 24 
$E\left(\frac{24}{7},\frac{24}{7}\right)$  4 × $\frac{24}{7}$ + 3 × $\frac{24}{7}$ = $\frac{196}{7}$= 24 
$B\left(0,6\right)$  4 × 0 + 3 × 6 = 18 
We see that the maximum value of the objective function Z is 24 which is at $F\left(5,\frac{4}{3}\right)$ and $E\left(\frac{24}{7},\frac{24}{7}\right)$.
Thus, the optimal value of Z is 24.
As, we know that if a LPP has two optimal solution, then there are an infinite number of optimal solutions.
Therefore, the given objective function can be subjected at an infinite number of points.
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Question 10:
If the constraints in a linear programming problem are changed
(a) the problem is to be reevaluated
(b) solution is not defined
(c) the objective function has to be modified
(d) the change in constraints is ignored
Answer:
(a) the problem is to be reevaluated
The optimisation of the objective function of a LPP is governed by the constraints.
Therefore, if the constraints in a linear programming problem are changed, then the problem needs to be reevaluated.
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Question 11:
Which of the following statements is correct?
(a) Every LPP admits an optimal solution
(b) A LPP admits unique optimal solution
(c) If a LPP admits two optimal solution it has an infinite number of optimal solutions
(d) The set of all feasible solutions of a LPP is not a converse set
Answer:
(c) If a LPP admits two optimal solutions, it has an infinite number of optimal solutions
It is known that the optimal solution of an LPP eiher exists uniquely, does not exist or exists infinitely.
So, If a LPP admits two optimal solution it has an infinite number of optimal solutions
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Question 12:
Which of the following is not a convex set?
(a) {(x, y) : 2x + 5y < 7}
(b) {(x, y) : x^{2} + y^{2} ≤ 4}
(c) {x :x = 5}
(d) {(x, y) : 3x^{2} + 2y^{2} ≤ 6}
Answer:
(c) {x :x = 5}
x = 5 is not a convex set as any two points from negative and positive xaxis if are joined will not lie in set.
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Question 13:
By graphical method, the solution of linear programming problem
$\mathrm{Maximize}Z=3{x}_{1}+5{x}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{Subject}\mathrm{to}3{x}_{1}+2{x}_{2}\le 18\phantom{\rule{0ex}{0ex}}{x}_{1}\le 4\phantom{\rule{0ex}{0ex}}{x}_{2}\le 6\phantom{\rule{0ex}{0ex}}{x}_{1}\ge 0,{x}_{2}\ge 0,\mathrm{is}$
(a) x_{1} = 2, x_{2} = 0, Z = 6
(b) x_{1} = 2, x_{2} = 6, Z = 36
(c) x_{1} = 4, x_{2} = 3, Z = 27
(d) x_{1} = 4, x_{2} = 6, Z = 42
Answer:
(b) x_{1} = 2, x_{2} = 6, Z = 36
We need to maximize the function Z = 3x_{1} + 5x_{2}
First, we will convert the given inequations into equations, we obtain the following equations:
3x_{1} + 2x_{2} = 18, x_{1} = 4, x_{2} = 6, x_{1} = 0 and x_{2} = 0
Region represented by 3x_{1} + 2x_{2} ≤ 18:
The line 3x_{1} + 2x_{2}_{ }= 18 meets the coordinate axes at A(6, 0) and B(0, 9) respectively. By joining these points we obtain the line 3x_{1} + 2x_{2}_{ }= 18.
Clearly (0,0) satisfies the inequation 3x_{1} + 2x_{2}_{ }= 18 .So,the region in the plane which contain the origin represents the solution set of the inequation
3x_{1} + 2x_{2}_{ }= 18.
Region represented by x_{1} ≤ 4:
The line x_{1} = 4 is the line that passes through C(4, 0) and is parallel to the Y axis. The region to the left of the line x_{1} = 4 will satisfy the inequation x_{1} ≤ 4.
Region represented by x_{2} ≤ 6:
The line x_{2} = 6 is the line that passes through D(0, 6) and is parallel to the X axis. The region below the line x_{2}_{ }= 6 will satisfy the inequation x_{2} ≤ 6.
Region represented by x_{1} ≥ 0 and x_{2} ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x_{1} ≥ 0 and x_{2} ≥ 0.
The feasible region determined by the system of constraints, 3x_{1} + 2x_{2} ≤ 18, x_{1} ≤ 4, x_{2} ≤ 6, x_{1} ≥ 0, and x_{2} ≥ 0, are as follows
Corner points are O(0, 0), D(0, 6), F(2, 6), E(4, 3) and C(4, 0).
The values of the objective function at these points are given in the following table
Points  Value of Z 
O(0, 0)  3(0)+5(0) = 0 
D(0, 6)  3(0)+5(6) = 30 
F(2, 6)  3(2)+5(6) = 36 
E(4, 3)  3(4)+5(3) = 27 
C(4, 0)  3(4)+5(0) = 12 
We see that the maximum value of the objective function Z is 36 which is at F(2, 6).
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Question 14:
The region represented by the inequation system x, y ≥ 0, y ≤ 6, x + y ≤ 3 is
(a) unbounded in first quadrant
(b) unbounded in first and second quadrants
(c) bounded in first quadrant
(d) none of these
Answer:
(c) bounded in first quadrant
Converting the given inequations into equations, we obtain
$y=6,x+y=3,x=0\mathrm{and}y=0$
y = 6 is the line passing through (0, 6) and parallel to the X axis.The region below the line y = 6 will satisfy the given inequation.
The line x + y = 3 meets the coordinate axis at A(3, 0) and B(0, 3). Join these points to obtain the line x + y =3.
Clearly, (0, 0) satisfies the inequation x + y ≤ 3. So, the region in xyplane that contains the origin represents the solution set of the given equation.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The shaded region represents the feasible region of the given LPP, which is bounded in the first quadrant
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Question 15:
The point at which the maximum value of x + y, subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x, y ≥ 0 is obtained, is
(a) (30, 25)
(b) (20, 35)
(c) (35, 20)
(d) (40, 15)
Answer:
(d) (40, 15)
We need to maximize the function Z = x + y
Converting the given inequations into equations, we obtain
$x+2y=70,2x+y=95,x=0\mathrm{and}y=0$
Region represented by x + 2y ≤ 70:
The line x + 2y = 70 meets the coordinate axes at A(70, 0) and B(0, 35) respectively. By joining these points we obtain the line
x + 2y = 70. Clearly (0,0) satisfies the inequation x + 2y ≤ 70. So,the region containing the origin represents the solution set of the inequation x + 2y ≤ 70.
Region represented by 2x + y ≤ 95:
The line 2x + y = 95 meets the coordinate axes at $C\left(\frac{95}{2},0\right)$ and D(0,95) respectively. By joining these points we obtain the line 2x + y = 95.
Clearly (0,0) satisfies the inequation 2x + y ≤ 95. So,the region containing the origin represents the solution set of the inequation
2x + y ≤ 95.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are O(0, 0), $C\left(\frac{95}{2},0\right)$ ,$E\left(40,15\right)$ and B(0, 35).
The values of Z at these corner points are as follows.
Corner point  Z = x + y 
O(0, 0)  0 + 0 = 0 
$C\left(\frac{95}{2},0\right)$  $\frac{95}{2}$+0 =$\frac{95}{2}$ 
$E\left(40,15\right)$  40 +15 = 55 
B(0, 35)  0 + 35 = 35 
We see that the maximum value of the objective function Z is 55 which is at $\left(40,15\right)$.
Page No 30.68:
Question 16:
The value of objective function is maximum under linear constraints
(a) at the centre of feasible region
(b) at (0, 0)
(c) at any vertex of feasible region
(d) the vertex which is maximum distance from (0, 0)
Answer:
(c) at any vertex of feasible region
In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value.
Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.
Page No 30.68:
Question 17:
The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
(a) p = q
(b) p = 2q
(c) p = 3q
(d) q = 3p
Answer:
(d) q = 3p
The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points (3, 4) and (0, 5).
∴ Value of Z at (3, 4) = Value of Z at (0, 5)
⇒ p(3) + q(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ q = 3p
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