A cottage industry manufactures pedestal lamps and wooden shades each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs. 5 and that from a shade is Rs. 3. Assuming that the manufacturer can sell ail the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit ?

.: Let the cottage industry manufacture x pedestal lamps and y wooden shades.

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

Since the grinding/cutting machine is available for atmost 12 hours

and sprayer is available for atmost 20 hours, we have the constraints
$2x + y \le 12$

$3x + 2y \le 20$

Total profit earned is $Z = 5x + 3y$

Hence, the mathematical formulation of the problem is Maximize $Z = 5x + 3y$ ...(1)

subject to the constraints
$2x + y \le 12$ ...(2)

$3x + 2y \le 20$ ...(3)

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

let ${l_{1l}}:2x + y = 12,{l_2}:3x + 2y = 20$

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

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
A(6, 0), E(4, 4) and B(0, 10).

We find that maximum value of Z is 32 at E(4, 4).

Hence, the manufacturer should sell 4 lamps and 4 shades to

realise a maximum profit and maximum profit is Rs. 32.