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.
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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