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The problem of maximizing Z = x1 – x2 subject to constrains x1 + x2 ≤ 10, x1 ≥ 0, x2 ≥ 0 and x2 ≤ 5 has
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- no solution
- one solution
- two solutions
- more than two solutions
Correct Option: B
max z = x1 – x2, such that:
x1 + x2 ≤ 10
x2 ≤ 5
x1 ≥ 0,x2 ≥ 0
Now, z (0, 0) = 0
z (0, 5) = 0 – 5 = – 5
z (10, 0) = 10 – 0 = 10
z (5, 5) = 5 – 5 = 0
Thus, max z occurs at only one point (10, 0) and the maximum value of z is 10 which is unique optimal solution.
