Question

A factory makes two types of items A and B, made of plywood. One piece of item A requires 5 minutes for cutting and 10 minutes for assembling. One piece of item B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours and 20 minutes available for cutting and 4 hours for assembling. The profit on one piece of item A is Rs 5 and that on item B is Rs 6. How many pieces of each type should the factory make so as to maximise profit? Make it as an L.P.P. and solve it graphically.

Answer 1

Let the factory makes x pieces of item A and B by pieces of item. 

Time required by item A (one piece) 

cutting = 5 minutes, assembling = 10 minutes 

Time required by item B (one piece) 

cutting = 8 minutes, assembling = 8 minutes 

Total time cutting = 3 hours & 20 minutes, assembling = 4 hours 

Profit on one piece item A = Rs 5, item B = Rs 6 

Thus, our problem is maximized Z = 5x + 6y

Subject to x ≥ 0, y ≥ 0 

5x + 8y ≤ 200 

10x + 8y ≤ 240 

On plotting graph of above constraints or inequalities, we get shaded region.

From figure, possible points for maximum value of z are at (24, 0), (8, 20), (0, 25). at (24, 0), z = 120 

at (8, 20), z = 40 + 120 = 160 (maximum) 

at (0, 25), z = 150 

∴ 8 pieces of item A and 20 pieces of item B produce maximum profit of Rs 160.