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Math Help - Business Math problem help

  1. #1
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    Business Math problem help

    I've worked on this problem for a couple of hours and I think my problem is that I don't know where to start....

    Georgia Cabinets manufactures kitchen cabinets that are sold to local dealers throughout the Southeast. The company has a large backlog of orders for oak and cherry cabinets and has decided to contract with three smaller cabinetmakers to do the final finishing operation.
    For Cabinetmaker 1 it'll take them 50 hours to complete all the oak cabinets, 60 hours to complete all cherry cabinets, but they only have 40 hours available. It'll cost $36 per hour.
    For Cabinetmaker 2 it'll take them 42 hours to complete all the oak cabinets, 48 horus to complete all cherry cabinets, but they only have 30 hours available. It'll cost $42 per hour.
    For Cabinetmaker 3 it'll take them 30 hours to complete all the oak cabinets, 35 hours to complete all cherry cabinets, and they only have 35 hours available. It'll cost $55 per hour.
    ((So for example, since Cabinetmaker 1 estimates it'll take 50 hours to complete all the oak, but only have 40 hours; then 40 hours available/50 hours to complete the oak = they'll finish only 80% of the oak.))

    I need to formulate a linear programming model that cna be used to determine the percentage of oak cabinets and the percentage of cherry cabinets that should be given to each of the three cabinetmakers in order to minimize the total cost of completing both projects.


    Thanks, any and all hope is much appreciated.
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  2. #2
    Junior Member F.A.P's Avatar
    Joined
    Dec 2006
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    Let x1 and x2 be hours spent on oak and cherry respectively by Cabinetmaker 1
    Let x3 and x4 be hours spent on oak and cherry respectively by Cabinetmaker 2
    Let x5 and x6 be hours spent on oak and cherry respectively by Cabinetmaker 3

    Costs per hour for each cabinetmaker are $36, $42 and $55.
    Let C be the total cost of completing the projects. That is, you want to minimize

    C=36(x_1 + x_2)+42(x_3 + x_4)+55(x_5 + x_6)

    subject to

    x_1 + x_2\leq40
    x_3 + x_4\leq30
    x_5 + x_6\leq35

    \frac{x_1}{50}+\frac{x_3}{42}+\frac{x_5}{30}=1 (finishing % at each maker must ad up to 100% for oak cabinets)

    \frac{x_2}{60}+\frac{x_4}{48}+\frac{x_6}{36}=1 (finishing % at each maker must ad up to 100% for charry cabinets)

    x_1,x_2,x_3,x_4,x_5,x_6\geq0

    Some adjustments to the way these conditions are stated and one should
    find a solution using simplex.. if one exists.
    Last edited by F.A.P; December 20th 2006 at 03:48 AM.
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