Q. 355.0( 1 Vote )

A manufacture produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type required 3 hours on machine A and 2 hours on Machine A and 2 hours on machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of ₹30 and ₹25 per trunk of the first type and second type respectively. How may trunks of each type must he make each day to make the maximum profit?

Answer :

Let the manufacturer manufacture x and y numbers of type 1 and type 2trunks.

According to the question,

3X + 3y

Maximize Z = 30x + 25y

The feasible region determined 3X + 3y is given by

The corner points of feasible region are A(0,0) , B(0,6) , C(3,3) , D(5,0).The value of Z at corner point is

The maximum value of Z is165 and occurs at point (3,3).

The manufacturer should manufacture 3 trunks of each type to earn maximum profit of Rs.165.

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