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Question
Minimize Z = x − 5y + 20
Subject to
\[x - y \geq 0\]
\[ - x + 2y \geq 2\]
\[ x \geq 3\]
\[ y \leq 4\]
\[ x, y \geq 0\]
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Solution
First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 0, − x + 2y = 2, x = 3, y = 4, x = 0 and y = 0.
Region represented by x − y ≥ 0 or x ≥ y:
The line x − y = 0 or x = y passes through the origin.The region to the right of the line x = y will satisfy the given inequation.
Let's check by taking an example like if we take a point (4, 3) to the right of the line x= y .Here x ≥ y.So, it satisfy the given inequation. Take a point (4, 5) to the left of the line x = y. Here, x ≤ y. That means it does not satisfy the given inequation.
Region represented by − x + 2y ≥ 2:
The line − x + 2y = 2 meets the coordinate axes at A(−2, 0) and B(0, 1) respectively. By joining these points we obtain the line
− x + 2y = 2.Clearly (0,0) does not satisfies the inequation − x + 2y ≥ 2. So,the region in xy plane which does not contain the origin represents the solution set of the inequation − x + 2y ≥ 2 .
The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. x ≥ 3 is the region to the right of the line x = 3.
The line y = 4 is the line that passes through the point (0, 4) and is parallel to X axis. y ≤ 4 is the region below the line y = 4.
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 x − y ≥ 0,− x + 2y ≥ 2, x ≥ 3, y ≤ 4, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are \[C\left( 3, \frac{5}{2} \right)\], \[D\left( 3, 3 \right)\], E(4, 4) and F(6, 4).
The values of Z at these corner points are as follows.
| Corner point | Z = x − 5y + 20 | |
|
\[C\left( 3, \frac{5}{2} \right)\]
|
3 − 5 × \[\frac{5}{2}\]+20= \[\frac{21}{2}\]
|
|
|
\[D\left( 3, 3 \right)\]
|
3 − 5 × 3 + 20 = 8 | |
| E(4, 4) | 4 − 5 × 4 + 20 = 4 | |
| F(6, 4) |
|
Therefore, the minimum value of Z is 4 at the point E(4, 4). Hence, x = 4 and y = 4 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 4.
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