Question

Maximize Z = 9x + 3y
Subject to
$$\begin{gathered} 2x+3y\le 13 \\ 3x+y\le 5 \\ x,y\ge 0 \end{gathered}$$

Answer 1

First, we will convert the given inequations into equations, we obtain the following equations:
2x + 3y = 13, 3x +y = 5, x = 0 and y = 0

Region represented by 2x + 3y ≤ 13 :
The line 2x + 3y = 13 meets the coordinate axes at $A\left(\frac{13}{2},0\right)$ and $B\left(0,\frac{13}{3}\right)$ respectively. By joining these points we obtain the line 2x + 3y = 13.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 13. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 13.

Region represented by 3x + y ≤ 5:
The line 5x + 2y = 10 meets the coordinate axes at $C\left(\frac{5}{3},0\right)$ and D(0, 5) respectively. By joining these points we obtain the line 3x + y = 5.
Clearly (0,0) satisfies the inequation 3x + y ≤ 5. So,the region containing the origin represents the solution set of the inequation 3x + y ≤ 5.

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 + 3y ≤ 13, 3x + y ≤ 5, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O(0, 0), $C\left(\frac{5}{3},0\right)$ ,$E\left(\frac{2}{7},\frac{29}{7}\right)$ and $B\left(0,\frac{13}{3}\right)$.

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

Corner point	Z = 9x + 3y
O(0, 0)	9 × 0 + 3 × 0 = 0
$C\left(\frac{5}{3},0\right)$	9 × $\frac{5}{3}$ + 3 × 0 = 15
$E\left(\frac{2}{7},\frac{29}{7}\right)$	9 × $\frac{2}{7}$ + 3 × $\frac{29}{7}$ = 15
$B\left(0,\frac{13}{3}\right)$	9 × 0 + 3 ×$\frac{13}{3}$ = 13

We see that the maximum value of the objective function Z is 15 which is at C $\left(\frac{5}{3},0\right)$ and $E\left(\frac{2}{7},\frac{29}{7}\right)$.
Thus, the optimal value of Z is 15.