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Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
  • a)
    Maximum Z = 12 at (2, 6)
  • b)
    Z has no maximum value
  • c)
    Maximum Z = 10 at (2, 6)
  • d)
    Maximum Z = 14 at (2, 6)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, ...
Objective function is Z = - x + 2 y ……………………(1).
The given constraints are : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Corner points Z =  - x + 2y

Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.
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Most Upvoted Answer
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, ...
Solution:

The given problem can be represented as follows:

Maximize Z = x + 2y

Subject to the constraints:

x ≥ 3
x – y ≥ 5
x – 2y ≥ 6
y ≥ 0

Graphical method:

We can solve the problem graphically by plotting the feasible region on a graph and finding the maximum value of Z. However, upon plotting the feasible region, we find that it is unbounded, i.e., it extends infinitely in all directions. Therefore, the objective function Z has no maximum value.

Explanation:

The feasible region for this problem is represented by the shaded region in the following graph:

The feasible region is unbounded, as it extends infinitely in all directions. Therefore, the objective function Z has no maximum value.

Conclusion:

Hence, the correct answer is option 'B'.
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Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.a)Maximum Z = 12 at (2, 6)b)Z has no maximum valuec)Maximum Z = 10 at (2, 6)d)Maximum Z = 14 at (2, 6)Correct answer is option 'B'. Can you explain this answer?
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