Question

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 the 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. 25 and that from a shade is Rs. 15. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit. Formulate an LPP and solve it graphically.

Answer 1

Let the manufacturer produces x padestal lamps and y wooden shades; then time taken by x pedestal lamps and y wooden shades on grinding/cutting machines = (2x + y) hours and time taken on the sprayer = (3x + 2y) hours.

Since grinding/cutting machine is available for at the most 12 hours.

∴ 2x + y ≤ 12

and sprayer is available for at most 20 hours. Thus, we have

∴ 3x + 2y ≤ 20

Now profit on the sale of x lamps and y shades is,

Z = 25x + 15y.

So, our problem is to find x and y so as to

Maximise Z = 25x + 15y …(i)

Subject to the constraints:

3x + 2y ≤ 20 …(ii)

2x + y ≤ 12 …(iii)

x ≥ 0 …(iv)

y ≥ 0 …(v)

The feasible region (shaded) OABC determined by the linear inequalities (ii) to (v) is shown in the figure. The feasible region is bounded.

Let us evaluate the objective function at each corner point as shown below:

| Corner points | Z = 25x + 15y |
| O(0, 0) | 0 |
| A(6, 0) | 150 |
| B(4, 4) | 160 Maximum |
| C(0, 10) | 150 |

We find that maximum value of Z is Rs. 160 at B(4, 4). Hence, manufacturer should produce 4 lamps and 4 shades to get maximum profit of Rs. 160.