Q. 29

A firm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. They need certain nutrients, named as X,Y,Z. the pigs are fed on two products, A and B. One unit of product A contain 36 unit of X, 3 units of Y and 20 units of Z, while one unit of product B contain 6 units of X, 12 units of Y and 10 units of Z. the minimum requirement of X, Y, Z are 108 units, 36 units and 100 units respectively. Product A costs ₹20 per unit and product B costs ₹40 per unit. How many units of each product must be taken to minimize the cost? Also, find the minimum cost.

Answer :

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

According to the question,

36x + 6y , 3x + 12y

Minimize Z = 20x + 40y

The feasible region determined 36x + 6y , 3x + 12y is given by

The feasible region is unbounded. The corner points of feasible region are A(0,18) , B(2,6) , C(4,2) , D(12,0).The value of Z at corner points are

The minimum value of Z is 160 at point (4,2).

Hence, the firm should buy 4 units of fertilizer A and 2 units of fertilizer B to achieve minimum expense of Rs.160.

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