Question

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 Rs17.50 per package on nuts and Rs7.00 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 the most 12 hours a day?

Answer 1

Let the manufacturer produce x packages of nuts and y packages of bolts.

Therefore, x≥0 and y≥0 The given information can be compiled in a table as follows.

| Nuts| Bolts| Availability
---|---|---|---
Machine(A)h| 1| 3| 12
Machine(B)h| 3| 1| 12

The profit of packaging of nuts is Rs17.50 and on the packaging of bolts is Rs7.

Therefore, the constraints are x+3y≤12 3x+y≤12 Total profit,Z=17.5x+7y

The mathematical formulation of the given problem is
Maximise Z=17.5x+7y.....(1)
Subject to the constraints,
x+3y≤12...(2)
x,y≥0...(4)

The feasible region determined by the system of constraints is as follows.

The corner points are A(4,0),B(3,3),C(0,4).
The values of Z at these corner points are as follows.

The maximum value of Z is Rs.73.50 at(3,3).

Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs.73.50.