Question

A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:

	Area occupied by the machine	Labour force for each machine	Daily output in units
Machine A Machine B	1000 sq. m 1200 sq. m	12 men 8 men	60 40

He has an area of 7600 sq. m available and 72 skilled men who can operate the machines.
How many machines of each type should he buy to maximize the daily output?

Answer 1

Let x machines of type A and y machines of type B were purchased.
Number of machines cannot be negative.
Therefore, $x,y\ge 0$

We are given,

	Area occupied by the machine	Labour force for each machine	Daily output in units
Machine A Machine B	1000 sq. m 1200 sq. m	12 men 8 men	60 40

The area of 7600 sq m is available and there are 72 skilled men available to operate the machines.

Therefore, the constraints are

$$\begin{gathered} 1000x+1200y\le 7600 \\ \mathrm{and}12x+8y\le 72 \end{gathered}$$

Total daily output = Z = $60x+40y$ which is to be maximised.

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

Max Z = $60x+40y$

subject to

$$\begin{gathered} 1000x+1200y\le 7600 \\ 12x+8y\le 72 \end{gathered}$$
$x,y\ge 0$

First we will convert inequations into equations as follows :
1000x + 1200y = 7600, 12x + 8y = 72, x = 0 and y = 0

Region represented by 1000x + 1200y ≤ 7600:
The line 1000x + 1200y = 7600 meets the coordinate axes at $A_1\left(\frac{38}{5},0\right)$ and $B_1\left(0,\frac{19}{3}\right)$ respectively. By joining these points we obtain the line
1000x + 1200y = 7600. Clearly (0,0) satisfies the 1000x + 1200y = 7600. So, the region which contains the origin represents the solution set of the inequation 1000x + 1200y ≤ 7600.

Region represented by 12x + 8y ≤ 72:
The line 12x + 8y = 72 meets the coordinate axes at C₁(6, 0) and D₁(0, 9) respectively. By joining these points we obtain the line 12x + 8y = 72 .Clearly (0,0) satisfies the inequation 12x + 8y ≤ 72. So,the region which contains the origin represents the solution set of the inequation 12x + 8y ≤ 72.


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 1000x + 1200y ≤ 7600, 12x + 8y ≤ 72, x ≥ 0, and y ≥ 0 are as follows.



The corner points are O(0, 0) $B_1\left(0,\frac{19}{3}\right)$, E₁(4, 3), C₁(6, 0)

The values of Z at these corner points are as follows

Corner point	Z= 60x + 40y
O	0
B1	253.3
E1	360
C1	360

The maximum value of Z is 360 which is attained at E₁(4, 3) and C₁(6, 0).

Thus, the maximum output is Rs 360 obtained when 4 units of type A and 3 units of type B or 6 units of type A are manufactured.