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?<p></p>

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?

Solution:

Let x be the number of packages of screws A and y be the number of packages of screws B that we can make. Clearly, x, y ≥ 0. Total time on machine is 4 hr = 240 min. Therefore, for the automatic machine, time = 4x + 6y ≤ 240 ⇒ 2x + 3y ≤ 120. For the hand machine, time = 6x + 3y ≤ 240 ⇒ 2x + y ≤ 80. The shaded region is the feasible region. The profits on Screw A is Rs. 0.7 and on Screw B is Rs. 1. We need to maximize the profits, i.e. maximize z = 0.7x + y, given the above constraints. Now, at A, z = 28. At O, z = 0. At E, z = 41. At D, z = 40. Hence, maximum profit is at point E(30, 20).

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