# 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  ?

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 = 120Z = 4 × 5 + 3 × 30 = 20 + 90 = 110Z = 4 × 20 + 3 × 15 = 80 + 45 = 125Z = 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.

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