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Question
Maximize Z = 50x + 30y
Subject to
\[2x + y \leq 18\]
\[3x + 2y \leq 34\]
\[ x, y \geq 0\]
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Solution
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 18, 3x + 2y = 34
Region represented by 2x + y ≥ 18:
The line 2x + y = 18 meets the coordinate axes at A(9, 0) and B(0, 18) respectively. By joining these points we obtain the line 2x + y = 18.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 18. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 18.
Region represented by 3x + 2y ≤ 34:
The line 3x + 2y = 34 meets the coordinate axes at
By joining these points we obtain the line 3x + 2y= 34.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 34. So,the region containing the origin
represents the solution set of the inequation 3x + 2y ≤ 34.
The corner points of the feasible region are A(9, 0),
The values of Z at these corner points are as follows.
| Corner point | Z = 50x + 30y | |
| A(9, 0) | 50 × 9 + 3 × 0 = 450 | |
| \[C\left( \frac{34}{3}, 0 \right)\] | 50 x \[\frac{34}{3}\] + 30 x 0= \[\frac{1700}{3}\] | |
| E(2, 14) |
|
Therefore, the maximum value of Z is
Thus, the optimal value of Z is \[\frac{1700}{3}\] .
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