A factory manufactures two types of screws A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs. 7 and screws 5 at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit ? Determine the maximum profit.

. : Let the manufacturer produces x packages of screws A and y packages of screws B.

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

Since the machine is available for at the most 4 hours a day,

we have the constraints
$4x + 6y \le 240$

$6x + 3y \le 240$

Total profit Z on package of screws A and 5 is $Z = 7x + 10y$

Hence, the mathematical formulation of the problem is

Maximize $Z = 7x + 10y$ ...(1)
subject to the constraints

$2x + 2y \le 120$ ...(2)

$2x + y \le 80$ ...(3)

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

Let : ${l_1}:4x + 6y = 240,{l_2} = 6x + 3y = 240$

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

The feasible region is shown in the graph.

Here again, we observe that the region is bounded.

Let us evaluate Z at the comer points
C(40, 0), E(30, 20) and B(0, 40).

We find that the maximum value of Z is 410 at E(30, 20).

Hence, the factory must sell 30 packages of screws and 20 packages of screw B

to realise maximum profit and maximum profit is Rs 410.