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Math Help - Linear Programming using graphs

  1. #1
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    Linear Programming using graphs

    Suppose that a plastic products manufacturer has to decide the weekly production levels of two types of lampshades:Colonial & Italiano. They are produced in packs of a dozen each. There are constraints on the amount of plastic available(500kg) and processing time (40 hrs) each week.

    Also, there should at most 700 dozen packages produced in total, and customer demand requires that the number of packs of Colonial lampshades should not exceed the number of packs of Italiano lampshades by more than 350.

    The resource & profit data ia as follows:


    Colonial $8.00 (profit per pack) 1 (plastic kg per pack) 3(production time minutes per pack)

    Italiano $5.00 0.5 4

    Formulate this as a LP to find the production levels that maximise the profit, and solve it using the graphical method.
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  2. #2
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    Quote Originally Posted by tondie2 View Post
    Suppose that a plastic products manufacturer has to decide the weekly production levels of two types of lampshades:Colonial & Italiano. They are produced in packs of a dozen each. There are constraints on the amount of plastic available(500kg) and processing time (40 hrs) each week.

    Also, there should at most 700 dozen packages produced in total, and customer demand requires that the number of packs of Colonial lampshades should not exceed the number of packs of Italiano lampshades by more than 350.

    The resource & profit data ia as follows:


    Colonial $8.00 (profit per pack) 1 (plastic kg per pack) 3(production time minutes per pack)

    Italiano $5.00 0.5 4

    Formulate this as a LP to find the production levels that maximise the profit, and solve it using the graphical method.

    The decision variables:
    Let
    C = number of packs of Colonial to be produced per week,
    I = number of packs of Italiano to be produced per week.

    -----------------------------
    The constraints:

    there should at most 700 dozen packages produced in total,
    So,
    C +I <= 700(12)
    C +I <= 8400 ------------(1)

    the number of packs of Colonial lampshades should not exceed the number of packs of Italiano lampshades by more than 350.
    So,
    C -I <= 350 ---------------(2)

    ...the amount of plastic available(500kg)
    ....Colonial, 1 plastic kg per pack
    ....Italiano, 0.5 plastic kg per pack

    so,
    C(1) +I(0.5) <= 500
    2C +I <= 1000 -----------------(3)

    ...and processing time (40 hrs)
    ...Colonial, 3 production minutes per pack
    ...Italiano, 4 production minutes per pack

    I assume "production minutes" is same as "processing minutes" here, so,
    C(3/60) +I(4/60) <= 40
    3C +4I <= 2400 ----------------(4)

    The non-negative constraints,
    C >= 0 -------------(5)
    I >= 0 -------------(6)

    I don't know how to graph here so graph the 6 constraints on paper. Say, horizontal axis is for C, so vertical axis is for I.
    Let me assume you know how to do that.
    Let me assume also that you know how to get the intersection of two inequalities.

    The inequality C +I <= 8400 ------------(1) is way out of the feasible region, rendering it useless. But the feasible region is in the correct side of the said inequality.
    I suspect then that it should have been for "at most 700 packages produced in total", not for 700 dozens packages as posted.

    The feasible region is a pentagon whose 5 corner points, in the order (C,I), are:
    (0,0) ---------intersection of (5) and (6)
    (0,600) -------intersection of (6) and (4)
    (320,360) -----intersection of (3) and (4)
    (450,100) -----intersection of (3) and (2)
    (350,0) -------intersection of (2) and (5)

    ----------------------------------------------
    Objective: to maximize profit.

    Objective function:
    P = C(8.00) +I(5.00)
    P = 8C +5I ---------------(i)

    Test that on all 5 vertices:
    (0,0) ======> P = 0
    (0,600) ====> P = 8*0 +5*600 = 3000
    (320,360) ==> P = 8*320 +5*360 = 4360
    (450,100) ==> P = 8*450 +5*100 = 4100
    (350,0) ====> P = 8*350 +5*0 = 2800

    Therefore, maximum profit ($4360 per week) is when 320 packs of Colonial lampshades and 360 packs of Italiano lampshades are produced per week. -------answer.
    Last edited by ticbol; August 29th 2007 at 04:24 AM.
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