Q. 165.0( 1 Vote )

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 availble 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 manufactrures, 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.[CBSE 2018]

Answer :

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

 x≥0, y≥0

Given that

The constraints are

4x + 6y ≤ 240

6x + 3y ≤ 240

Total profit: Z = 0.70x + y

Subject to,

2x + 3y ≤ 120

2x + y ≤ 120

x≥0, y≥0.

Now, plotting the equations 2x + 3y ≤ 120 and 2x + y ≤ 120 we get,


the common feasible region is OCBAO.

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 get the maximum profit of Rs.41.

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