A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs. 17.50 per package on nuts and Rs. 7 per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at most 12 hours a day? Formulate the above as a linear programming problem and solve it graphically.
Answer: Max profit = Rs 73.50 at (3, 3): 3 packages of nuts and 3 packages of bolts
Solution
Let x packages of nuts and y packages of bolts be produced per day. Profit: Z=17.5x+7y. Machine A: x+3y≤12 Machine B: 3x+y≤12 x≥0,y≥0 Corner points: (0,0), (4,0), (3,3) (intersection of x+3y=12 and 3x+y=12) and (0,4). Evaluate Z=17.5x+7y: (4,0): Z=70 (3,3): Z=52.5+21=73.5 ← maximum (0,4): Z=28
∴Zmax=Rs 73.50 at (3,3)
3 packages of nuts and 3 packages of bolts should be produced each day.Feasible region for x + 3y = 12 and 3x + y = 12, with corner points (0, 0), (4, 0), (3, 3) and (0, 4).
