Question

Maximize Z = 9x + 3y
Subject to 

2x + 3y ≤ 13

3x + y ≤ 5

x, y ≥ 0

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.

We see that the maximum value of the objective function Z is 15 which is at C

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 × \[\frac{13}{3}\] = 113

\[\left( \frac{5}{3}, 0 \right)\] and  \[E\left( \frac{2}{7}, \frac{29}{7} \right)\]  Thus, the optimal value of Z is 15.