Question

Maximize Z = 15x + 10y
Subject to
$$\begin{gathered} 3x+2y\le 80 \\ 2x+3y\le 70 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0

Region represented by 3x + 2y ≤ 80 :
The line 3x + 2y = 80 meets the coordinate axes at $A\left(\frac{80}{3},0\right)$ and $B\left(0,40\right)$ respectively. By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 80 . So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 80 .

Region represented by 2x + 3y ≤ 70:
The line 2x + 3y = 70 meets the coordinate axes at $C\left(35,0\right)$ and $D\left(0,\frac{70}{3}\right)$ respectively. By joining these points we obtain the line 2x + 3y ≤ 70.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 70. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 70.

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, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), $A\left(\frac{80}{3},0\right)$ ,$E\left(20,10\right)$ and $D\left(0,\frac{70}{3}\right)$.

The values of Z at these corner points are as follows.

Corner point	Z = 15x + 10y
O(0, 0)	15 × 0 + 10 × 0 = 0
$A\left(\frac{80}{3},0\right)$	15 × $\frac{80}{3}$ + 10 × 0 = 400
$E\left(20,10\right)$	15 × 20 + 10 × 10 = 400
$D\left(0,\frac{70}{3}\right)$	15 × 0 + 10 ×$\frac{70}{3}$ = $\frac{700}{3}$

We see that the maximum value of the objective function Z is 400 which is at $A\left(\frac{80}{3},0\right)$ and $E\left(20,10\right)$.
Thus, the optimal value of Z is 400.