Question

A manufacturing company makes two models A and B of a product. Each piece of Model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs 9000 on each piece of model A and Rs 11000 on each piece of Model B.How many pieces of Model A and Model B should be manufactured per week to realise a maximum profit?

• A. 20 pieces of model A & 0 pieces of model B
• B. 0 pieces of model A & 10 pieces of model B
• C. 12 pieces of model A & 6 pieces of model B
• D. 10 pieces of model A & 10 pieces of model B

Answer 1

The correct option is A 20 pieces of model A & 0 pieces of model B

Suppose x is the number of pieces of Model A and y is the number of pieces
of Model B manufactured.
Then,
Total profit (in Rs) = 9000x+11000y
Let Z = 9000x+11000y
Maximise Z = 9000 x+11000y
subject to the constraints:
Fabricating constraint
9x + 12y ≤ 180
i.e. 3x + 4y ≤ 60
Finishing constraint
x + 3y ≤ 30
non-negative constraint
x ≥ 0, y ≥ 0
The feasible region OABC determined by the linear inequalities is shown in the figure. Note that the feasible region is bounded

Corner Point	Z = 9000x+11000y
A(0, 10)	110000
O(0, 0)	0
B(20, 0)	180000
C(12, 6)	174000

Hence, the company should produce 20 pieces of Model A and 0 pieces of Model B to realise maximum profit.