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Question
Maximize Z = 9x + 3y
Subject to
2x + 3y ≤ 13
3x + y ≤ 5
x, y ≥ 0
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Solution
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.
Region represented by 3x + y ≤ 5:
The line 5x + 2y = 10 meets the coordinate axes at
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),
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
|
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