Question

A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit.

Answer 1

Let x  units and y units of articles A and B are produced respectively.
Number of articles cannot be negative.
Therefore,  \[x, y \geq 0\]

The product of each unit of article A requires 4 hours in assembly and that of article B requires 2 hours in assembly and the maximum capacity of the assembly department is 60 hours a week 

\[4x + 2y \leq 60\]

The product of each unit of article A requires 2 hours in finishing and that of article B  requires 4 hours in assembly and the maximum capacity of the finishing department is 48 hours a week

\[2x + 4y \leq 48\]

If the profit is Rs 6 for each unit of A and Rs 8 for each unit of B. Therefore, profit gained from​ x units and y units of articles A and B respectively is Rs 6x and Rs 8y respectively.
Total revenue = Z =  \[6x + 8y\] which is to be maximised.
Thus, the mathematical formulat​ion of the given linear programmimg problem is

Max Z =  \[6x + 8y\]

subject to

\[2x + 4y \leq 48\]
\[4x + 2y \leq 60\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows:
2x + 4y = 48, 4x + 2y = 60, x = 0 and y = 0

Region represented by 2x + 4y ≤ 48:
The line 2x + 4y = 48 meets the coordinate axes at A₁(24, 0) and \[B_1 \left( 0, 12 \right)\]  respectively. By joining these points we obtain the line 2x + 4y = 48. Clearly (0,0) satisfies the 2x + 4y = 48. So, the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 48.
Region represented by 4x + 2y ≤ 60:
The line 4x + 2y = 60 meets the coordinate axes at C₁(15, 0) and  \[D_1 \left( 0, 30 \right)\] respectively. By joining these points we obtain the line 4x + 2y = 60. Clearly (0,0) satisfies the inequation 4x + 2y ≤ 60. So,the region which contains the origin represents the solution set of the inequation 4x + 2y ≤ 60.

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 2x + 4y ≤ 48, 4x + 2y ≤ 60, x ≥ 0 and y ≥ 0 are as follows.

 

T
he corner points are O(0, 0), B₁(0, 12), E₁(12, 6) and C₁(15,0). 
The values of Z at these corner points are as follows

Corner point	Z= 6x + 8y
O	0
B1	96
E1	120
C1	90

The maximum value of Z is 120 which is attained at E₁(12, 6).
Thus, the maximum profit is Rs 120 obtained when 12 units of article A and 6 units of article B were manufactured.