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Question : 29
Total: 29
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 ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture 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 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 factory manufacture x screws of type A and y screws of type B per day.
Hence , x ≥ 0 , y ≥ 0
According to the question, Z = 0.7x + y
4x + 6y ≤ 40
6x + 3y ≤ 240
The maximum value of Z is 41 at (30,20).
Hence, the factory should produce 30 packages of screws A
and 20 packages of screws B to get the maximum profit of Rs. 41.
Hence , x ≥ 0 , y ≥ 0
| Screw A | Screw B | Availiability | |
|---|---|---|---|
| Automatic Machine | 4 | 6 | 4 × 60 = 240 |
| Hand Operated Machine | 6 | 3 | 4 × 60 = 240 |
According to the question, Z = 0.7x + y
4x + 6y ≤ 40
6x + 3y ≤ 240
| Z = 0.7x + y | |
|---|---|
| P (40 , 0) | 28 |
| Q (30 , 20) | 41 |
| R (0 , 40) | 40 |
The maximum value of Z is 41 at (30,20).
Hence, the factory should produce 30 packages of screws A
and 20 packages of screws B to get the maximum profit of Rs. 41.
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