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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 h on grinding / cutting machine and 3 h on the sprayer to manufacture a pedestal lamp. It takes 1 h on the grinding / cutting machine and 2 h on the sprayer to Manufacture a shade. On any day, the sprayer is available for at the most 20 h and the grinding / cutting machine for at most 12 h. The profit from the sale of a lamp is Rs. 5 and that from a shades is Rs. 3. Assuming that the manufacture can sell all the lamps and shades that he produce, how should he schedule his daily production in order to maximize his profit?

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Solution

Let the manufacturer produces x pedestal lamps and y wooden shades everyday. We construct the following table :

ItemNumberTime on grindingTime onProfitcutting machine(in h)sprayer(in h)(in Rs.)Ax2x3x5xByy2y3yTotalx+y2x+y3x+2y5x+3yAvailability1220

The profit on a lamp is Rs. 5 and on the shades is Rs. 3

Our problem is to maximize Z = 5x + 3y .......(i)

Subject to the constraints 2x + y ≤ 12 ......(ii)

3x + 2y ≤ 20 .......(iii)

x≥0, y≥0 .....(iv)

Firstly, draw the graph of the line 2x + y = 12

x06y120

Putting (0, 0) in the inequality 2x+y≤12, we have 2×0+0≤12⇒0≤12 (which is true)

So, the half plane is towards the origin, Since, x,y≥0

So, the feasible region lies in the first quadrant. Secondly, draw the graph of the line 3x+2y=20

x0203=6.6y100

Putting (0, 0) in the inequality 3x+2y≤20, we have

3×0+2×0≤20⇒0≤20 (which is true)

So, the half plane is towards the origin.

On solving equations 2x+y=12 and 3x+2y=20, we get B(4, 4).

∴ Feasible region is OABCO.

The corner points of the feasible region are O(0, 0), A(6, 0), B(4, 4) and C(0, 10). The values of Z at these points are as follows :

Corner pointZ=5x+3y0(0, 0)0A(6, 0)30B(4, 4)32→MaximumC(0, 10)30

The maximum value of Z is Rs. 32 at B(4, 4).

Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.

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