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 machines 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 least 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 all the lamps and shades that the produces, how should he schedule his daily production in order to maximize his profit?

Answer 1

Hint: First of all, let the number of pedestal lamps that are to be made is $x$ and the number of wooden shades that are to be made be $y$. Then, use the given conditions to form the corresponding linear programming problem. Solve the problem by using the graph method. Find the value of the objective function at the corner points. Then, the value at which the profit is maximum is the optimal value.

Complete step by step Answer:

We will begin by letting the number of pedestal lamps that are to be made is $x$ and the number of wooden shades that are to be made be $y$.
We are given that the pedestal lamps require 2 hours of grinding/cutting and wooden shades require 1 hour of grinding/cutting and there are maximum 12 hours for cutting/grinding.
Then, we can say that $2x + y \leqslant 12$ (1)
Similarly, we are also given that the pedestal lamps require 3 hours of sprayer whereas the wooden shades require 2 hours for the sprayer. Also, the maximum number of hours available for sprayer is 20 hours.
Hence, $3x + 2y \leqslant 20$ (2)
Also, the number of pedestal lamps and wooden shades cannot be negative.
Therefore, $x \geqslant 0,y \geqslant 0$
We have to maximize profit.
We are given that the profit on a pedestal lamp is Rs. 5 and the profit on wooden shades is Rs.3.
Hence, we have to maximize $Z = 5x + 3y$
We will solve the linear programming problem:
Maximize $Z = 5x + 3y$
Subject to $2x + y \leqslant 12$
And $3x + 2y \leqslant 20$
$x \geqslant 0,y \geqslant 0$
We will first plot the linear inequalities and plot the boundary points.

The value of the function $Z = 5x + 3y$ at boundary points is:

Point	$Z = 5x + 3y$
P$\left( {0,10} \right)$	$5\left( 0 \right) + 3\left( {10} \right) = 30$
Q$\left( {4,4} \right)$	$5\left( 4 \right) + 3\left( 4 \right) = 32$
R\[\left( {6,0} \right)\]	$5\left( 6 \right) + 3\left( 0 \right) = 30$

The maximum value is at the point Q$\left( {4,4} \right)$
Hence, industry should produce 4 units of pedestal lamp and 4 units of wooden shades in order to maximise profit.

Note: We use linear programming method when we have limit resources and have to use most of it. In this type of questions, many students are unable to translate the description in terms of inequalities correctly. Here, we were the maximum of the resources and hence we used the less than equal to sign. Also, the shading of the area should be done correctly.