Maximize Z = 3x + 4y
Subject to
\[2x + 2y \leq 80\]
\[2x + 4y \leq 120\]
Solution
We have to maximize Z = 3x + 4y
First, we will convert the given inequations into equations, we obtain the following equations:
2x + 2y = 80, 2x + 4y = 120
Region represented by 2x + 2y ≤ 80:
The line 2x + 2y = 80 meets the coordinate axes at \[A\left( 40, 0 \right)\] and \[B\left( 0, 40 \right)\] respectively. By joining these points we obtain the line 2x + 2y = 80.
Clearly (0,0) satisfies the inequation 2x + 2y ≤ 80. So,the region containing the origin represents the solution set of the inequation 2x + 2y ≤ 80.
Region represented by 2x + 4y ≤ 120:
The line 2x + 4y = 120 meets the coordinate axes at
Clearly (0,0) satisfies the inequation 2x + 4y ≤ 120. So,the region containing the origin represents the solution set of the inequation 2x + 4y ≤ 120.
The feasible region determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:
Corner point  Z = 3x + 4y 
O(0, 0)  3 × 0 + 4 × 0 = 0 
\[A\left( 40, 0 \right)\]

3× 40 + 4 × 0 = 120 
\[E\left( 20, 20 \right)\]

3 × 20 + 4 × 20 = 140 
\[D\left( 0, 30 \right)\]

10 × 0 + 4 ×30 = 120 
We see that the maximum value of the objective function Z is 140 which is at \[E\left( 20, 20 \right)\] that means at x = 20 and y = 20.
Thus, the optimal value of Z is 140.