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Computing the Dual Prices for Optimal Linear Program Solution

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The optimal solution of the linear programming problem is at the intersection of constraints 1 and 2.
Max 2x1 + x2
s.t. 4x1 + 1x2 < 400
4x1 + 3x2 < 600
1x1 + 2x2 £ 300
x1 , x2 > 0

Compute the dual prices for the three constraints.
.45, .25, 0
.25, .25, 0
0, .25, .45
.45, .25, .25

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Solution Summary

Shows how to compute the dual prices in a few paragraphs' worth of explained calculations.

Solution Preview

Since constraint 3 is not a binding constraint the dual price for third constraint is zero. Thus, possible answers are .45, .25, 0 or .25, .25, 0.
The solution lies at the intersection of constraints 1 and 2. Thus, ...

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