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Linear Programming:
  A Word Problem with Four Variables
(page 5 of 5)

Sections: Optimizing linear systems, Setting up word problems


  • A building supply has two locations in town. The office receives orders from two customers, each requiring 3/4-inch plywood. Customer A needs fifty sheets and Customer B needs seventy sheets.
  • The warehouse on the east side of town has eighty sheets in stock; the west-side warehouse has forty-five sheets in stock. Delivery costs per sheet are as follows: $0.50 from the eastern warehouse to Customer A, $0.60 from the eastern warehouse to Customer B, $0.40 from the western warehouse to Customer A, and $0.55 from the western warehouse to Customer B.

    Find the shipping arrangement which minimizes costs.

    Hmm... I've got four things to consider:

      east warehouse to Customer A
      east warehouse to Customer B
      west warehouse to Customer A
      west warehouse to Customer B

    But I only have two variables. How can I handle this?

    The variables obviously need to stand for the number of sheets being shipped, but I have four different sets of sheets. This calls for subscripts and explicit labelling:

      shipped from east warehouse to Customer A: Ae
      shipped from west warehouse to Customer A:
      Aw
      shipped from east warehouse to Customer B:
      Be
      shipped from west warehouse to Customer B:
      Bw

    Since Customer A wants 50 sheets and Customer B wants 70 sheets, then:

      Ae + Aw = 50, so Aw = 50  Ae   (I'll call this Equation I)
      Be
      + Bw = 70,
      so Bw = 70  Be   (I'll call this Equation II)

    Since the eastern warehouse can ship no more than eighty sheets and the western warehouse can ship no more than forty-five sheets, then:

      0 < Ae + Be < 80
      0 < Aw + Bw < 45

    And the optimization equation will be the shipping cost:

      C = 0.5Ae + 0.6Be + 0.4Aw + 0.55Bw

    Where does this leave me? The equations (labelled as Equations I and II above) allow me to substitute and get rid of two of the variables in the second inequality above, so:

      0 < Ae + Be < 80
      0 < (50  Ae) + (70  Be) < 45

     

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    Simplifying the second inequality above gives me the new system:

      0 < Ae + Be < 80
      0 < 120  Ae  Be < 45

    Multiplying through by 1 (thereby flipping the inequality signs) and adding 120 to all three "sides" of the second inequality, I get the new system:

      0 < Ae + Be < 80
      120 > AeBe > 75

    Since Ae + Be is no less than 75 and is no more than 80, then these two inequalities reduce to one:

      75 < Ae + Be < 80

    I can also simplify the optimization equation:

      C = 0.5Ae + 0.6Be + 0.4Aw + 0.55Bw
          
      = 0.5Ae + 0.6Be + 0.4(50  Ae) + 0.55(70  Be)
          = 0.1Ae + 0.05Be + 58.50

    Due to the size of the orders, I have:

      0 < Ae < 50
      0 < Be < 70

    Since I have only two variables now, and since I'll be graphing with x and y, I'll rename the variables:   Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

      x = Ae
      y = Be

    Then entire system is as follows:

      Minimize C = 0.1x + 0.05y + 58.50, subject to the following:

        x > 0
        y > 0
        y > x + 75

        x < 50
        y < 70
        y < x + 80

    The feasibility region graphs as:

      feasibility region

When you test the corner points, (5, 70), (10, 70), (50, 30), and (50, 25), you should get the minimum cost when you ship as follows:

      5 sheets from the eastern warehouse to Customer A
    70 sheets from the eastern warehouse to Customer B
    45 sheets from the western warehouse to Customer A
      0 sheets from the western warehouse to Customer B

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Cite this article as:

Stapel, Elizabeth. "Linear Programming: A Word Problem with Four Variables."
    Purplemath. Available from
http://www.purplemath.com/modules/linprog5.htm.
    Accessed
 

 



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