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Consider the following linear programming problem (LPP).
Maximize z = 2x + 3y
Subject to
2x + 3y ≤ 12
2x - 3y ≤ 0
y ≤ 2
x, y ≥ 0
It has
- a)no feasible solution
- b)unique optimal solution
- c)alternative optimal solutions
- d)unbounded solution
Correct answer is option 'B'. Can you explain this answer?
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Consider the following linear programming problem (LPP).Maximize z = ...
Problem:
Maximize z = 2x - 3y
Subject to
2x - 3y ≤ 12
2x - 3y ≤ 0
y ≤ 2x
y ≥ 0
Feasible Region:
To find the feasible region, we need to graph the given inequalities and identify the region that satisfies all the constraints.
The first constraint, 2x - 3y ≤ 12, can be rewritten as y ≥ (2/3)x - 4. We can graph this line by plotting the points (0, -4) and (6, 0) and drawing a line through them.
The second constraint, 2x - 3y ≤ 0, can be rewritten as y ≥ (2/3)x. We can graph this line by plotting the points (0, 0) and (3, 2) and drawing a line through them.
The third constraint, y ≤ 2x, can be graphed by plotting the points (0, 0) and (1, 2) and drawing a line through them.
Finally, the constraint y ≥ 0 represents the non-negative region of the y-axis.
After graphing all the constraints, we can shade the region that satisfies all the constraints. The feasible region is the intersection of all the shaded regions.
Analysis:
In this case, the feasible region is a bounded region consisting of a triangle. Since the objective function is linear, it will have a maximum value at one of the corner points of the feasible region.
Corner Points:
To find the corner points, we need to find the intersections of the lines representing the constraints.
- Intersection of (2/3)x - 4 and (2/3)x: (6, 0)
- Intersection of (2/3)x - 4 and 2x: (3, 2)
- Intersection of (2/3)x and 2x: (0, 0)
Objective Function:
Now, we evaluate the objective function z = 2x - 3y at each of the corner points.
- At (6, 0): z = 2(6) - 3(0) = 12
- At (3, 2): z = 2(3) - 3(2) = 0
- At (0, 0): z = 2(0) - 3(0) = 0
Optimal Solution:
The maximum value of z is 12, which occurs at the corner point (6, 0). Therefore, the LPP has a unique optimal solution.
Conclusion:
The correct answer is option 'B', which states that the LPP has a unique optimal solution.
Maximize z = 2x - 3y
Subject to
2x - 3y ≤ 12
2x - 3y ≤ 0
y ≤ 2x
y ≥ 0
Feasible Region:
To find the feasible region, we need to graph the given inequalities and identify the region that satisfies all the constraints.
The first constraint, 2x - 3y ≤ 12, can be rewritten as y ≥ (2/3)x - 4. We can graph this line by plotting the points (0, -4) and (6, 0) and drawing a line through them.
The second constraint, 2x - 3y ≤ 0, can be rewritten as y ≥ (2/3)x. We can graph this line by plotting the points (0, 0) and (3, 2) and drawing a line through them.
The third constraint, y ≤ 2x, can be graphed by plotting the points (0, 0) and (1, 2) and drawing a line through them.
Finally, the constraint y ≥ 0 represents the non-negative region of the y-axis.
After graphing all the constraints, we can shade the region that satisfies all the constraints. The feasible region is the intersection of all the shaded regions.
Analysis:
In this case, the feasible region is a bounded region consisting of a triangle. Since the objective function is linear, it will have a maximum value at one of the corner points of the feasible region.
Corner Points:
To find the corner points, we need to find the intersections of the lines representing the constraints.
- Intersection of (2/3)x - 4 and (2/3)x: (6, 0)
- Intersection of (2/3)x - 4 and 2x: (3, 2)
- Intersection of (2/3)x and 2x: (0, 0)
Objective Function:
Now, we evaluate the objective function z = 2x - 3y at each of the corner points.
- At (6, 0): z = 2(6) - 3(0) = 12
- At (3, 2): z = 2(3) - 3(2) = 0
- At (0, 0): z = 2(0) - 3(0) = 0
Optimal Solution:
The maximum value of z is 12, which occurs at the corner point (6, 0). Therefore, the LPP has a unique optimal solution.
Conclusion:
The correct answer is option 'B', which states that the LPP has a unique optimal solution.
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Consider the following linear programming problem (LPP).Maximize z = 2x + 3ySubject to2x + 3y ≤ 122x - 3y ≤ 0y ≤ 2x, y ≥ 0It hasa)no feasible solutionb)unique optimal solutionc)alternative optimal solutionsd)unbounded solutionCorrect answer is option 'B'. Can you explain this answer?
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