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Maximise Z = − x + 2y, subject to the constraints:
.
Asked by Y Last Modified
Answer 
Vishaka Bhutia
Home Tutor
The feasible region determined by the constraints, 
is as follows.
It can be seen that the feasible region is unbounded.
The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows.
| Corner point | Z = −x + 2y |
| A(6, 0) | Z = − 6 |
| B(4, 1) | Z = − 2 |
| C(3, 2) | Z = 1 |
As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, −x + 2y > 1, and check whether the resulting half plane has points in common with the feasible region or not.
The resulting feasible region has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Z has no maximum value.
read less
It can be seen that the feasible region is unbounded.
The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows:
As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.
For this, we graph the inequality, −x + 2y > 1, and check whether the resulting half plane has
points in common with the feasible region or not.
The resulting feasible region has points in common with the feasible region.
Therefore, Z = 1 is not the maximum value. Z has no maximum value.
read less Manali Shah
Tutor
The feasible region determined by the constraints, x + 2y ≤ 120, x + y ≥ 60, x − 2y ≥ 0, x ≥ 0, and
y ≥ 0 is as follows:
The corner points of the feasible region are A (60, 0), B (120, 0), C (60, 30), and D (40, 20).
The values of Z at these corner points are as follows:
The minimum value of Z is 300 at (60, 0) and the maximum value of Z is 600 at all the points on
the line segment joining (120, 0) and (60, 30).
read lessView 1 more Answers
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