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.
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|
|B(30, 20)||41||→ Maximum|
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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