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Determine the best feasible solution for the below LP problem: min z= 4x1 3x2, 2x1 x2≥ 10,-3x1 2x2≤6,x1 x2≥ 6,x1 &x2 ≥ 0 A. (4,2) Ops: B. (2,4) c. (5,2) D. (2,5)?
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Determine the best feasible solution for the below LP problem: min z= ...
X2 ≥ 0
To solve the LP problem, we can use the graphical method by plotting the feasible region and finding the corner points.
First, we plot the constraints:
2x1 + x2 ≥ 10 (red line)
-3x1 + 2x2 ≤ 6 (blue line)
x1 x2 ≥ 6 (green line)
x1, x2 ≥ 0 (quadrant 1)
The feasible region is the shaded area:
The corner points of the feasible region are:
A: (2, 6)
B: (2, 3)
C: (3, 2)
D: (6, 1)
E: (6, 2)
Next, we evaluate the objective function z = 4x1 + 3x2 at each corner point:
A: z = 4(2) + 3(6) = 26
B: z = 4(2) + 3(3) = 14
C: z = 4(3) + 3(2) = 18
D: z = 4(6) + 3(1) = 27
E: z = 4(6) + 3(2) = 30
The smallest value of z is at point B: z = 14.
Therefore, the best feasible solution for the LP problem is x1 = 2, x2 = 3, with a minimum value of z = 14.
To solve the LP problem, we can use the graphical method by plotting the feasible region and finding the corner points.
First, we plot the constraints:
2x1 + x2 ≥ 10 (red line)
-3x1 + 2x2 ≤ 6 (blue line)
x1 x2 ≥ 6 (green line)
x1, x2 ≥ 0 (quadrant 1)
The feasible region is the shaded area:
The corner points of the feasible region are:
A: (2, 6)
B: (2, 3)
C: (3, 2)
D: (6, 1)
E: (6, 2)
Next, we evaluate the objective function z = 4x1 + 3x2 at each corner point:
A: z = 4(2) + 3(6) = 26
B: z = 4(2) + 3(3) = 14
C: z = 4(3) + 3(2) = 18
D: z = 4(6) + 3(1) = 27
E: z = 4(6) + 3(2) = 30
The smallest value of z is at point B: z = 14.
Therefore, the best feasible solution for the LP problem is x1 = 2, x2 = 3, with a minimum value of z = 14.
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Determine the best feasible solution for the below LP problem: min z= 4x1 3x2, 2x1 x2≥ 10,-3x1 2x2≤6,x1 x2≥ 6,x1 &x2 ≥ 0 A. (4,2) Ops: B. (2,4) c. (5,2) D. (2,5)?
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