Mathematics · Linear Programming
The objective function of the dual problem for the following primal problem is: Max(Z) = 3x1 - 3x2 s.t. x1 ≤ 4 x2 ≤ 6 x1 + x2 ≤ 5 -x2 ≤ -1 x1, x2 ≥ 0
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1Min(Z') = 4w1 + 6w2 + 5w3 + w4
- 2Min(Z') = -4w1 - 6w2 + 5w3 - w4
- 3Max(Z') = -4w1 - 6w2 - 5w3 + w4
- 4Min(Z') = 4w1 + 6w2 + 5w3 - w4✓
Correct answer
Min(Z') = 4w1 + 6w2 + 5w3 - w4
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