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Question
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A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand - operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of Rs. 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.

Solution

Let the factory manufactures x screws of type A and y screws of type B on each day.

∴ x ≥ 0, y ≥ 0

Given that

  Screw A Screw B Availability
Automatic Machine 4 6 4 x 60 = 240 minutes
Hand operate machine 6 3 4 x 60 = 240 minutes
Profit 70 paise 1 rupee  

The constraints are

4x + 6y ≤ 240

6 x + 3y ≤ 240

Total profit

z = 0.70 x + 1y

∴l.P.P is

maximise z = 0.7 x + y

subject to,

2x +3y ≤ 120

2x + y ≤ 80

x ≥0, y ≥0

∴ common feasible region is OCBAO

Correct point Z = 0.7x + y
A (40,0) Z(A) = 28
B (30,20) Z (B) = 41 maximum
C (0,40) Z(C) = 40
O(0,0) Z(O) = 0

The maximum value of 'Z' is 41 at (30,20). Thus the factory showed produce 30 packages at screw A and 20 packages of screw B to ge the maximum profit of Rs.41

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