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Math Help - Linear Programming - Sensitivity Range of One Constraint

  1. #1
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    Exclamation Linear Programming - Sensitivity Range of One Constraint

    Hey guys, I've attached the work I have done... I'm not exactly sure why I am wrong, or what appears to be wrong with my math.


    The Question: imgur: the simple image sharer
    My Answer: imgur: the simple image sharer
    The correct answer is C1 = 58.33. C2 = 25.71


    The first part of the question is easy, A is obviously optimal. However the second question, as easy as it appears, is tricking me. Finding C1 and C2


    The textbook has taught me to rearranged the objective function in order to find its slope, which is -30/50 if I am finding C1 and -50/30 if I am finding C2. I also did the same for constraint line (x1/x2) which is -60/70 for C1 and then -70/60 for C2 (x2/x1) to reflect the same setup as the objective function.


    The way they go about doing it is -C1/50 = -70/60 and then -30/C2 = -70/60.

    I am totally lost because the textbook has taught me to do what I have done in my answer... Yet it's wrong. Can someone tell me what I am failing to realize? I am having trouble conceptualizing the problem and usually do math based on pattern recognition.

    Thanks for your time (apologize for potato quality pictures)
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  2. #2
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    Re: Linear Programming - Sensitivity Range of One Constraint

    To solve for c_{1}, we want the tangency condition: c_{1} * 60 = c_{2} * 70. Holding c_{2} = 50, we get c_{1} = \dfrac{350}{60} = 58.33.

    To solve for c_{2}, we hold c_{1} = 30 and solve: 30 * 60 = 70 * c_{2} to get c_{2} = \dfrac{1800}{70} = 25.71.
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