A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while 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 most 20 hours and the grinding/cutting machine for at most 12 hours. The profit from the sale of a lamp is ₹5.00 and a shade is ₹3.00. Assuming that the manufacturer sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? [NCERT, CBSE 2013]

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Solution

Suppose x units of pedestal lamps and y units of wooden shades are produced on a day to maximise the profit.

Since a pedestal lamp requires 2 hours on the grinding/cutting machine and a wooden shade requires 1 hour on the grinding/cutting machine, therefore, the total hours required for grinding/cutting x units of pedestal lamps and y units of wooden shades are (2x + y). But, the grinding/cutting machine is available for at most 12 hours on a day.

∴ 2x + y ≤ 12

Similarly, a pedestal lamp requires 3 hours on the sprayer and a wooden shade requires 2 hours on the sprayer, therefore, the total hours required for spraying x units of pedestal lamps and y units of wooden shades are (3x + 2y). But, the sprayer is available for at most 20 hours on a day.

∴ 3x + 2y ≤ 20

The profit from the sale of a pedestal lamp is ₹5.00 and a wooden shade is ₹3.00. Therefore, the total profit from the sale of x units of pedestal lamps and y units of wooden shades is ₹(5x + 3y).

Thus, the given linear programming problem is

Maximise Z = 5x + 3y

subject to the constraints

2x + y ≤ 12

3x + 2y ≤ 20

x, y ≥ 0

The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are O(0, 0), A(6, 0), B(4, 4) and C(0, 10).

The value of the objective function at these points are given in the following table.

Corner Point Z = 5x + 3y

(0, 0) 5 × 0 + 3 × 0 = 0

(6, 0) 5 × 6 + 3 × 0 = 30

(4, 4) 5 × 4 + 3 × 4 = 32 → Maximum

(0, 10) 5 × 0 + 3 × 10 = 30

The maximum value of Z is 32 at x = 4, y = 4.

Hence, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profit. The maximum profit of the manufacturer is ₹32 on a day.

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