Question

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 min each for cutting and 10 min each for assembling. Souvenirs of type B require 8 min each for cutting and 8 min each for assembling. There are 3 h 20 min available for cutting and 4 h for assembling. The profit is Rs. 5 each for type A and Rs. 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

Answer 1

Let the company manufactures x souvenirs of type A and y souvenirs of type B. We construct the following table :

$\begin{array}{ccccc}Type& Number& Timeforcutting& Timefor& Profit\\ & & machine\left(inmin\right)& assembling\left(inmin\right)& (inRs.)\\ A& x& 5x& 10x& 5x\\ B& y& 8y& 8y& 6y\\ Total& x+y& 5x+8y& 10x+8y& 5x+6y\\ Availability& & 3\times 60+20=200& 4\times 60=240& \end{array}$

The profit on type A souvenirs is Rs. 5 and on type B souvenirs is Rs. 6.

Our problem is to maximize Z = 5x + 6y ...........(i)

Subject to the constraints

$5x+8y\le 200$ .........(ii)

$10x+8y\le 240\iff 5x+4y\le 120$ ......(iii)

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

Firstly, draw the graph of the line $5x+8y=200$

$\begin{array}{ccc}x& 0& 40\\ y& 25& 0\end{array}$

Putting (0, 0) in the inequality $5x+8y\le 200$, we have

$5\times 0+8\times 0\le 200\Rightarrow 0\le 200$ (which is true)

So, the half plane is towards the origin.

Since, $x,y\ge 0$

So, the feasible region lies in the first quadrant

Secondly, draw the graph of the line $5x+4y=120$

$\begin{array}{ccc}x& 0& 24\\ y& 30& 0\end{array}$

Putting (0, 0) in the inequality $5x+4y\le 120$, we have

$5\times 0+4\times 0\le 120$

$\Rightarrow 0\le 120$ (which is true)

So, the half plane is towards the origin.

On solving equations $5x+8y=200$ and $5x+4y=120$, we get B(8, 20).

$\therefore$ Feasible region is OABCO

The corner points of the feasible region are O(0, 0), A(24, 0), B(8, 20) and C(0, 25). The values of Z at these points are as follows :

$\begin{array}{cc}Cornerpoint& Z=5x+6y\\ O(0,0)& 0\\ A(24,0)& 120\\ B(8,20)& 160\to Maximum\\ C(0,25)& 150\end{array}$

The maximum value of Z is Rs. 160 at B(8, 20).

Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs. 160.