Question

A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, A and B, are available at a cost of ₹4 and ₹3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.

Answer 1

Let x and y be number of units of food A and B.

∴According to the question,

200x + 100y ≥ 4000, x + 2y ≥ 50, 40x + 40y ≥ 1400, x ≥ 0, y ≥ 0

Minimize Z = 4x + 3y

The feasible region determined 200x + 100y ≥ 4000, x + 2y ≥ 50, 40x + 40y ≥ 1400, x ≥ 0, y ≥ 0 is given by

The feasible region is unbounded. The corner points of feasible region are A(0,40) , B(5,30) , C(20,15) , D(50,0).

The value of Z at corner points are

Corner Point	Z = 4x + 3y
A(0, 40)	120
B(5, 30)	110	Minimum
C(20, 15)	125
D(50, 0)	200

The minimum value of Z is 110 at point (5,30).

Hence, the diet should contain 5 units of food A and 30 units of food B for the least cost.