Question

A manufacturer has three machines installed in his factory. 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. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:

Item	Number of hours required by the machine
A B	I	II	III
	1 2	2 1	1 5/4

He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.

Answer 1

Let x units of item A and y units of item B be manufactured.
Therefore, $x,y\ge 0$
As we are given,

Item	Number of hours required by the machine
A B	I	II	III
	1 2	2 1	1 5/4

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

He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B.
Profit made by him in producing x items of A and y items of B 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 A₁(12, 0) and B₁(0, 6) 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 C₁(6, 0) and D₁(0, 12) 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 E₁(5, 0) and F₁(0, 4) 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 contains 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 B₁(0, 6), G₁(4, 4), C₁(6, 0), E₁(5, 0) and F₁(0, 4).

The values of Z at these corner points are as follows

Corner point	Z = 6x + 4y
B1	24
G1	40
C1	36
E1	30
F1	16

The maximum value of Z is 40 which is attained at G₁(4, 4).

Thus, the maximum profit is Rs 40 obtained when 4 units each of item A and B are manufactured.