Question

A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes $4$ minutes on the automatic and $6$ minutes on the hand-operated machines to manufacture a packet of screws 'A' while it takes $6$ minutes on the automatic and $3$ minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most $4$ hours on any day. The manufacturer can sell a packet of screws 'A' at a profit of $70$ paise and screws 'B' at a profit of Rs $1$. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.

Answer 1

Let $x$ and $y$ be the number of packets that can be manufactured of screws type '$A$' and type '$B$' respectively.

Objective function $Z$ is :

$Z\left(Maximise\right)=0.7x+y$

Subject to the constraints :

$4x+6y\le 240$ or $2x+3y\le 120$

$6x+3y\le 240$ or $2x+y\le 80$

And $x,y\ge 0$.

Consider $2x+3y=120$

Table of solutions is :

$\begin{array}{ccc}x& 0& 60\\ y& 40& 0\end{array}$

Consider $2x+y=80$

Table of solutions is :

$\begin{array}{ccc}x& 0& 40\\ y& 80& 0\end{array}$

Plot the points $A(40,0),B(0,80),C(0,40)$ and $D(60,0)$ on the same graph.

Join them to get the shaded region (bounded) as shown in the figure. The corner points of the shaded region are

$A(40,0),E(30,20),C(0,40)$ and $O(0,0)$.



At $A(40,0),Z=0.7\times 40+0=28$

At $E(30,20),Z=0.7\times 30+20$

= $41\leftarrow$ Maximum

At $C(0,40),Z=0+40=40$

At $O(0,0),Z=0+0=0$

Hence, a maximum profit of Rs 41 can be obtained by manufacturing 30 packets of type 'A' and 20 packets of type 'B' screws.