Question

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 units of item M and y units of item N be manufactured.
Therefore, $x,y\ge 0$

As we are given,

Item	Number of hours required by the machine
M N	I	II	III
	1 2	2 1	1 1.25

Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day.

According to question, the constraints are
$$\begin{gathered} x+2y\le 12 \\ 2x+y\le 12 \\ x+1.25y\ge 5\mathrm{or}x+\frac{5}{4}y\ge 5 \end{gathered}$$

He makes a profit of Rs 600 on item M and Rs 400 on item N.
Profit made by him in producing x items of M and y items of N is 6x + 4y.

Total profit Z = $6x+4y$ which is to be maximised

Thus, the mathematical formulat​ion of the given linear programmimg problem is

Max Z = $6x+4y$

subject to

$$\begin{gathered} x+2y\le 12 \\ 2x+y\le 12 \\ x+\frac{5}{4}y\ge 5 \end{gathered}$$
$x,y\ge 0$

First we will convert inequations into equations as follows :
x + 2y = 12, 2x + y = 12, $x+\frac{5}{4}y=5$, x = 0 and y = 0

Region represented by x + 2y ≤ 12:
The line x + 2y = 12 meets the coordinate axes at $A\left(12,0\right)$ and $B\left(0,6\right)$ respectively. By joining these points we obtain the line
x + 2y = 12.Clearly (0,0) satisfies the x + 2y = 12. So,the region which contains the origin represents the solution set of the inequation
x + 2y ≤ 12.

Region represented by 2x + y ≤ 12:
The line 2x + y = 12 meets the coordinate axes at $C\left(6,0\right)$ and $D\left(0,12\right)$ respectively. By joining these points we obtain the line
2x + y = 12.Clearly (0,0) satisfies the inequation 2x + y ≤ 12. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 12.

Region represented by $x+\frac{5}{4}y\ge 5$:
The line $x+\frac{5}{4}y=5$ meets the coordinate axes at $E\left(5,0\right)$ and $F\left(0,4\right)$ respectively. By joining these points we obtain the line
$x+\frac{5}{4}y=5$.Clearly (0,0) does not satisfies the inequation $x+\frac{5}{4}y\ge 5$. So,the region which does not contain the origin represents the solution set of the inequation $x+\frac{5}{4}y\ge 5$.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints x + 2y ≤ 12, 2x + y ≤ 12, $x+\frac{5}{4}y\ge 5$, x ≥ 0, and y ≥ 0 are as follows.





The corner points are B(0, 6), $G\left(4,4\right)$, C(6, 0), E(5, 0) and F(0, 4).

The values of Z at these corner points are as follows

Corner point	Z = 600x + 400y
B	2400
G	4000
C	3600
E	3000
F	1600

The maximum value of Z is 4000 which is attained at G(4, 4).

Thus, the maximum profit is Rs 4000 obtained when 4 units each of item M and N are manufactured.