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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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    https://brainmass.com/math/linear-programming/computing-dual-prices-optimal-linear-program-solution-143741

    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, ...

    Solution Summary

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

    $2.49

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