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. 7.00 per packag

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 Rs.17.50 per package on nuts and Rs. 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at most 12 hours a day ?

Accepted Answer

.: Let x be the number of package of nuts and y be the number of package of bolts produced by machine A and B.

Clearly $x \ge 0,y \ge 0.$ Now, we make the following table from the given data.

$figure$

Since the machine is operated for at most 12 hours a day, we have the constraints:

$x + 3y \le 12$

$3x + y \le 12$
Total profit Z earned on package of nuts and bolts is

$Z = 17.5x + 7y$
Hence, the mathematical formulation of the problem is

Maximize $Z = 17.5x + 7y$ ...(1)
subject to the constraints

$x + 3y \le 12$ ...(2)

$3x + y \le 12$ ...(3)

$x,y \ge 0$ ...(4)

Let ${l_1}:x + 3y = 12,{l_2}:3x + y = 12$

Let us graph the inequalities (2) to (4).

$figure$

The feasible region determined by the system is shown in the graph.

Here again, observe that the region is bounded.

Let us evaluate Z at the comer points
C(4, 0), E(3, 3) and B(0, 4).

We find that maximum value of Z is 73.5 at 5(3, 3).

Hence, the factory should produce 3 packages of nuts and 3 packages

of bolts to realise maximum profit and maximum profit then will be Rs. 73.50

Linear Programming — Class 12 Maths Solution