Question

A factory makes two types of items $A$ and $B$, made of plywood. One piece of item A requires $5$ minutes for cutting and $10$ minutes for assembling. One piece of item $B$ requires $8$ minutes for cutting and $8$ minutes for assembling. There are $3$ hours and $20$ minutes available for cutting and $4$ hours for assembling. The profit on one piece of item $A$ is Rs $5$ and that on item $B$ is Rs $6$. How many pieces of each type should the factory make so as to maximise profit? Make it as an L.P.P. and solve it graphically

Answer 1

Let the factory makes $x$ pieces of item $A$ and $B$ by pieces of item.

Time required by item $A$ (one piece)

cutting $=5$ minutes, assembling $=10$ minutes

Time required by item $B$ (one piece)

cutting $=8$ minutes, assembling $=8$ minutes

Total time

cutting $=3$ hours and $20$ minutes,

assembling $=4$ hours

Profit on one piece item $A=$ Rs $5$, item $B=$Rs $6$

Thus, our problem is maximized .

$Z=5x+6y$

Subject to $x\ge 0,y\ge 0$

$5x+8y\le 200$

$10x+8y\le 240$

From figure, possible points for maximum value of $z$ are at $(24,0)$, $(8,20)$,$(0,25)$.

At $(24,0),z=120$

at $(8,20)$,

$z=40+120=160$ (maximum)

at $(0,25)$, $z=150$

$8$ pieces of item $A$ and $20$ pieces of item $B$ produce maximum profit of Rs $160$.