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Linear Programming
Question
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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. 8000 on each piece of model A and Rs. 12000 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? What is the maximum profit per week?

Answer

Let x be the number of pieces of Model A and y be the number of pieces of Model B. Then
Total profit = Rs. (8000 x + 12000 y)
Let    Z = 8000 x + 12000 y
The mathematical formulation of the problem is as followings:
Maximise Z = 8000 x + 12000 y subject to constraints 9x + 12y ≤ 180
i.e. 3x + 4 y ≤ 60
x + 3y ≤ 30
x ≥ 0, y ≤ 0
Now we draw the graph of 3x + 4y = 60
For x = 0, 4y = 60 or y = 15
For y = 0, 3x = 60 or x = 20
∴  line meets OX in A (20, 0) and OY in B(0, 15).
Also we draw the graph of x + 3y = 30

For x = 0, 3 y = 30 or y = 10
For y = 0, x = 30
∴ line meets OX in B(30, 0) and OY in M(0, 10)
Since feasible region satisfies all the constraints.
∴  OACM is the feasible region.
The comer points are O(0, 0), A(20, 0), C(12, 6), M(0, 10).
At O(0, 0), Z = 0 + 0 = 0
At A(20, 0), Z = 160000 + 0 = 160000
At C(12, 6), Z = 96000 + 72000 = 168000
At M(0, 10), Z = 0 + 120000 = 120000
∴ maximum value = 168000 at (12, 6)
∴ company should produce 12 pieces of model A and 6 pieces of model B to realise maximum profit of Rs. 168000.

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