Question

Maximize Z = 3x + 4y
Subject to
$$\begin{gathered} 2x+2y\le 80 \\ 2x+4y\le 120 \end{gathered}$$

Answer 1

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 $C\left(60,0\right)$ and $D\left(0,30\right)$ respectively. By joining these points we obtain the line 2x + 4y ≤ 120.
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:

The corner points of the feasible region are O(0, 0), $A\left(40,0\right)$, $E\left(20,20\right)$ and $D\left(0,30\right)$.

The values of Z at these corner points 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.