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  4. A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for atmost 12 h whereas machine III must be operated for atleast 5 h 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 <th colspan=3>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

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Q. A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for atmost $12 h$ whereas machine III must be operated for atleast $5 h$ 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

Linear Programming

Solution:

Let $x$ and $y$ be the number of items $M$ and $N$, respectively.
Total profit on the production $=₹(600 x+400 y)$
Mathematical formulation of the given problem is as follows
Maximise $Z=600 x=400 y$
Subject to the constraints are
$x+2 y \leq 12 \text { (constraint on machine I) }$...(i)
$2 x+y \leq 12 \text { (constraint on machine II) } $...(ii)
$x+\frac{5}{4} y \geq 5 \text { (constraint on machine III) } $...(iii)
$x \geq 0, y \geq 0$...(iv)
Let us draw the graph of constraints (i) to (iv). $A B C D E$ 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.
image
Let us evaluate $Z=600 x+400 y$ at these corner points.
Corner point $Z=600 x+400 y$
$(5,0)$ 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 $₹ 4000$.