Given the Linear Programming Problem:
Maximize \( z = 11x + 7y \) subject to the constraints: \( x \leq 3 \), \( y \leq 2 \), \( x, y \geq 0 \).
Then the optimal solution of the problem is:
Given the Linear Programming Problem:
Maximize \( z = 11x + 7y \) subject to the constraints: \( x \leq 3 \), \( y \leq 2 \), \( x, y \geq 0 \).
Then the optimal solution of the problem is:
Show Hint
- \( (3, 2) \)
- \( (3, 0) \)
- \( (0, 2) \)
- \( (1, 0) \)
- \( (0, 1) \)
The Correct Option is A
Solution and Explanation
To solve this Linear Programming problem, we first graph the constraints \( x \leq 3 \), \( y \leq 2 \), and \( x, y \geq 0 \).
These constraints define the feasible region, which is a quadrilateral with vertices at \( (0, 0) \), \( (3, 0) \), \( (0, 2) \), and \( (3, 2) \).
Now, evaluate the objective function \( z = 11x + 7y \) at each of the vertices of the feasible region: - At \( (0, 0) \), \( z = 11(0) + 7(0) = 0 \)
- At \( (3, 0) \), \( z = 11(3) + 7(0) = 33 \)
- At \( (0, 2) \), \( z = 11(0) + 7(2) = 14 \)
- At \( (3, 2) \), \( z = 11(3) + 7(2) = 33 + 14 = 47 \) The maximum value of \( z \) is 47, which occurs at the point \( (3, 2) \).
Thus, the optimal solution is \( (3, 2) \).
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