Search

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

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:

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

<< Previous  Top  |  1 | 2 | 3 | 4 | 5  |  Return to Index

 Cite this article as: Stapel, Elizabeth. "Linear Programming: A Word Problem with Four Variables."     Purplemath. Available from https://www.purplemath.com/modules/linprog5.htm.     Accessed [Date] [Month] 2016

Reviews of
Internet Sites:
Free Help
Practice
Et Cetera

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor