A diet for a sick person must contain at least 4000 units of vitamins , 50 units of minerals , & 1400 of calories . Two foods A & B , are available at a cost of Rs, 4 & Rs , 3 per unit respec. If one unit of A containa 200 units of vitamin , 1 unit of mineral & 40 calories & one unit of food B contains 100 units of vitamin , 2 units of minerals & 40 calories , find what combination of foods should be used to have the least cost ?
Here is the answer to your question.
The given data may be put in the following tabular form:
| Food | Vitamins | Minerals | Calories | Cost per unit |
| A | 200 | 1 | 40 | Rs 4 |
| B | 100 | 2 | 40 | Rs 3 |
| Minimum requirement | 4000 | 50 | 1400 |
Suppose the diet contains x units of food A and y units of food B. Since, one unit of food A costs Rs 4 and one unit of food B costs Rs 3. Therefore, total cost of x units of food A and y units of food B is Rs (4x + 3y).
Let Z denotes the total cost.
Z = 4x + 3y
Since each unit of food A contains 200 units of vitamins. Therefore, x units of food A contain 200 x units of vitamins. A unit of food B contains 100 units of vitamins. Therefore, y units of food B contain 100 y units of vitamins.
Thus, x units of food A and y units of food B contain 200 x + 100 y units of vitamins. But, the minimum requirement of vitamins is 4000 units.
∴ 200 x + 100 y ≥ 4000
Similarly, the total amount of minerals supplied by x units of food A and y units of food B is x + 2y and the minimum requirement is of 50 units.
∴ x + 2y ≥50
Finally, the total calories in x units of food A and y units of food B is 40 x + 40 y and the minimum calories required is 1400 units.
∴ 40 x + 40 y ≥ 1400
Clearly, x ≥ 0 and y ≥ 0
Since, we have to minimize the total cost Z. Thus, the mathematical form of the given LPP is as follows:
Minimize Z = 4x + 3y
Subject to
200x + 100y ≥ 4000
x + 2y ≥ 50
40x + 40y ≥ 1400
x, y ≥ 0
The inequalities can now be represented on graph as:
The solution set of the linear constraints is shaded in the figure. The vertices of the shaded region are A (0, 40), P1 (5, 30), P2 (20, 15) and B (50, 0).
The value of the objective function at these points are given in the following table.
| Point (x, y) | Value of objective function Z = 4x + 3y |
A (0, 40) P1 (5, 30) P2 (20, 15) B (50, 0) | Z = 4 × 0 + 3 × 40 = 0 + 120 = 120 Z = 4 × 5 + 3 × 30 = 20 + 90 = 110 Z = 4 × 20 + 3 × 15 = 80 + 45 = 125 Z = 4 × 50 + 3 × 0 = 200 + 0 = 200 |
Clearly, Z is minimum for x = 5 and y = 30 and the minimum value of Z is 110.
Hence, the diet cost is minimum when 5 units of food A and 30 units of food B are taken.
The minimum diet cost is Rs 110.