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Question
Minimize Z = 5x + 3y
Subject to
\[2x + y \geq 10\]
\[x + 3y \geq 15\]
\[ x \leq 10\]
\[ y \leq 8\]
\[ x, y \geq 0\]
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Solution
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 10, x + 3y = 15, x = 10, y = 8
Region represented by 2x + y ≥ 10:
The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively. By joining these points we obtain the line 2x + y = 10.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 10. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y≥ 10.
Region represented by x + 3y ≥ 15:
The line x + 3y = 15 meets the coordinate axes at C(15, 0) and D(0, 5) respectively. By joining these points we obtain the line x + 3y = 15.
Clearly (0,0) satisfies the inequation x + 3y ≥ 15. o,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 3y ≥ 15.
The line x = 10 is the line that passes through the point (10, 0) and is parallel to Yaxis.x ≤ 10 is the region to the left of the line x = 10.
The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis.y ≤ 8 is the region below the line y = 8.
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 + y ≥ 10, x + 3y ≥ 15, x ≤ 10, y ≤ 8, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are E(3, 4), \[H\left( 10, \frac{5}{3} \right)\] F(10, 8) and G(1, 8).
| Corner point | Z = 5x + 3y |
| E(3, 4) | 5 × 3 + 3 × 4 = 27 |
|
\[H\left( 10, \frac{5}{3} \right)\]
|
5 × 10 + 3× \[\frac{5}{3}\] = 55
|
| F(10, 8) | 5 × 10 + 3 × 8 = 74 |
| G(1, 8) | 5 × 1 + 3 × 8 = 29 |
Thus, the optimal value of Z is 27.
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