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Question
Minimize Z = 30x + 20y
Subject to
\[x + y \leq 8\]
\[ x + 4y \geq 12\]
\[5x + 8y = 20\]
\[ x, y \geq 0\]
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Solution
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 0 and y = 0
5x + 8y = 20 is already an equation.
Region represented by x + y ≤ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.Clearly (0,0) satisfies the inequation x+ y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x + y ≤ 8.
Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.Clearly (0,0) satisfies the inequation x + 4y ≥ 12. So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.
The line 5x + 8y = 20 is the line that passes through E(4, 0) and \[F\left( 0, \frac{5}{2} \right)\] 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 ≤ 8, x + 4y ≥ 12, 5x+ 8y = 20, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are B(0,8), D(0,3),
The values of Z at these corner points are as follows.
| Corner point | Z = 30x + 20y |
| B(0,8) | 160 |
| D(0,3) | 60 |
|
\[G\left( \frac{20}{3}, \frac{4}{3} \right)\]
|
266.66 |
Therefore, the minimum value of Z is 60 at the point D(0,3). Hence, x = 0 and y =3 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 60
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