Question

A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of ₹50 each on type A and ₹60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?

Answer 1

Let the company manufacture x and y numbers of toys A and B.

∴According to the question,

5X + 8y ≤ 180, 10x + 8y ≤ 240, x ≥ 0, y ≥ 0

Maximize Z = 50x + 60y

The feasible region determined 5X + 8y ≤ 180, 10x + 8y ≤ 240, x ≥ 0, y ≥ 0 is given by

The corner points of feasible region are A(0,0) , B(0,22.5) , C(12,15) , D(24,0).

The value of Z at corner point is

Corner Point	Z = 50x + 60y
A(0, 0)	0
B(0, 22.5)	1350
C(12, 15)	1500	Maximum
D(24, 0)	1200

The maximum value of Z is1500 and occurs at point (12,15).

The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500.