Question

A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for atmost $12h$ whereas machine III must be operated for atleast $5h$ 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 ₹ 600 and ₹ 400 on items $M$ and $N$, respectively.
Then, to maximise the profit, number of units of item $M$, the manufacturer has to produce, is

• A. 4
• B. 5
• C. 6
• D. 3

Answer 1

Answer: (A)

Let $x$ and $y$ be the number of items $M$ and $N$, respectively.
Total profit on the production $=₹(600x+400y)$
Mathematical formulation of the given problem is as follows
Maximise $Z=600x=400y$
Subject to the constraints are
$x+2y\le 12\text{ (constraint on machine I) }$...(i)
$2x+y\le 12\text{ (constraint on machine II) }$...(ii)
$x+\frac{5}{4}y\ge 5\text{ (constraint on machine III) }$...(iii)
$x\ge 0,y\ge 0$...(iv)
Let us draw the graph of constraints (i) to (iv). $ABCDE$ is the feasible region (shaded) as shown in figure determined by the constraints (i) to (iv). 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	$Z=600x+400y$
$(5,0)$	300
$(6,0)$	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 $₹4000$.