Question
Z = 20x1 + 202, subject to x1 ≥ 0, x2 ≥ 0, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20. The minimum value of Z occurs at
Options :
(8, 0)
(5/2, 15/4)
(7/2, 9/4)
(0, 10)
Answer :
(7/2, 9/4)
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