Question

Maximize Z = 15x + 10y
Subject to 

\[3x + 2y \leq 80\]
\[2x + 3y \leq 70\]
\[ x, y \geq 0\]

 

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.