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Question
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 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 manufacture x screws of type A and y screws of type B on each day. Therefore,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.
| Screw A | Screw B | Availability | |
| Automatic Machine (min) | 4 | 6 | 4x60 = 240 |
| Hand Operated Machine (min) | 6 | 3 | 4x60 = 240 |
The profit on a package of screws A is 70 paise and on the package of screws, B is Rs 1. Therefore, the constraints are
`4x + 6y <= 240`
`6x + 3y < =- 240`
Total profit, Z = 0.7x + y
The mathematical formulation of the given problem is
Maximize Z = 0.7x + y … (1)
subject to the constraints,
`4x + 6y <= 240` ...2
`6x + 3y <= 240` ...3
x,y >= 0 ....4
The feasible region determined by the system of constraints is
The corner points are A (40, 0), B (30, 20), and C (0, 40).
The values of Z at these corner points are as follows.
| Corner point | Z = 0.7x + y | |
| A(40, 0) | 28 | |
| B(30, 20) | 41 | → Maximum |
| C(0, 40) | 40 |
The maximum value of Z is 41 at (30, 20).
Thus, 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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