Question

Consider the following linear programming problem:

Min Z = 10x1 + 6x2
Subject to:
4x1 + 3x2 ≥ 200
2x1 + x2 ≥ 70
x1, x2 ≥ 0
Listed are the 3 potential optimal solutions
A=(0,70)
B=(5,60)
C=(50,0)

Answer 1

Explanation:

To determine the optimal solution among the three potential solutions A, B, and C, we need to evaluate the objective function value (Z) for each solution and select the one with the minimum value.

Solution A = (0, 70):

Z_A = 10(0) + 6(70) = 420

Solution B = (5, 60):

Z_B = 10(5) + 6(60) = 370

Solution C = (50, 0):

Z_C = 10(50) + 6(0) = 500

Comparing the objective function values, we find that the minimum value is achieved at Solution B, which has an objective function value of 370. Therefore, the optimal solution is B = (5, 60) with Z = 370.