Question

(Manufacturing problem) A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most $12$ hours whereas machine III must be operated for atleast $5$ hours a day. She produces only two items $M$ and $N$ each requiring the use of all the three machines.
The number of hours required for producing $1$ unit of each of $M$ and $N$ on the three machines are given in the following table:

Items	Number of hours required on machines
	I	II	III
$M$	$1$	$2$	$1$
$N$	$2$	$1$	$1.25$

She makes a profit of $Rs.600$ and $Rs.400$ on items $M$ and $N$ respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?

Answer 1

Let $x$ and $y$ be the number of items $M$ and $N$ respectively.

Total profit on the production $=Rs(600x+400y)$
Mathematical formulation of the given problem is as follows :

Maximise $Z=600x+400y$

subject to the constraints :
$x+2y\le 12$ (constraint on Machine I) $...\left(1\right)$
$2x+y\le 12$ (constraint on Machine II) $...\left(2\right)$
$x+54y\ge 5$ (constraint on Machine III) $...\left(3\right)$
$x\ge 0,y\ge 0...\left(4\right)$

Let us draw the graph of constraints $\left(1\right)$ to $\left(4\right).ABCDE$ is the feasible region (shaded) as shown in Fig determined by the constraints $\left(1\right)$ to $\left(4\right)$. Observe that the feasible region is bounded, coordinates of the corner points $A,B,C,D$ and $E$ are $(5,0)(6,0),(4,4),(0,6)$ and $(0,4)$ respectively.

Let us evaluate $Z=600x+400y$ at these corner points.

Corner point	$Z=600x+400y$
$(5,0)$	$3000$
$(6,0)$	$3600$
$(4,4)$	$4000\leftarrow Maximum$
$(0,6)$	$2400$
$(0,4)$	$1600$

We see that the point $(4,4)$ is giving the maximum value of $Z$. Hence, the manufacturer has to produce $4$ units of each item to get the maximum profit of $Rs.4000$.