Linear programming is the process of taking various linear inequalities relating to some situation, and finding the "best" value obtainable under those conditions. A typical example would be taking the limitations of materials and labor, and then determining the "best" production levels for maximal profits under those conditions.

In "real life", linear programming is part of a very important area of mathematics called "optimization techniques". This field of study (or at least the applied results of it) are used every day in the organization and allocation of resources. These "real life" systems can have dozens or hundreds of variables, or more. In algebra, though, you'll only work with the simple (and graphable) two-variable linear case.

The general process for solving linear-programming exercises is to graph the inequalities (called the "constraints") to form a walled-off area on the x,y-plane (called the "feasibility region"). Then you figure out the coordinates of the corners of this feasibility region (that is, you find the intersection points of the various pairs of lines), and test these corner points in the formula (called the "optimization equation") for which you're trying to find the highest or lowest value.

1. Find the maximal and minimal value of z = 3x + 4y subject to the following constraints:

The three inequalities in the curly braces are the constraints. The area of the plane that they mark off will be the feasibility region. The formula "z = 3x + 4y" is the optimization equation. I need to find the(x, y) corner points of the feasibility region that return the largest and smallest values of z.

My first step is to solve each inequality for the more-easily graphed equivalent forms:

It's easy to graph the system:   Copyright © Elizabeth Stel 2006-2011 All Rights Reserved

To find the corner points -- which aren't always clear from the graph -- I'll pair the lines (thus forming a system of linear equations) and solve:

y = –( 1/2 )x + 7y = 3x

y = –( 1/2 )x + 7y = x – 2

y = 3xy = x – 2

–( 1/2 )x + 7 = 3x–x + 14 = 6x

14 = 7x2 = x

y = 3(2) = 6

–( 1/2 )x + 7 = x – 2–x + 14 = 2x – 4

18 = 3x6 = x

y = (6) – 2 = 4

3x = x – 22x = –2x = –1

y = 3(–1) = –3

corner point at (2, 6) corner point at (6, 4) corner pt. at (–1, –3)

So the corner points are (2, 6), (6, 4), and (–1, –3).

Somebody really smart proved that, for linear systems like this, the maximum and minimum values of the optimization equation will always be on the corners of the feasibility region. So, to find the solution to this exercise, I only need to plug these three points into "z = 3x + 4y".

(2, 6):      z = 3(2)   + 4(6)   =   6 + 24 =   30 (6, 4):      z = 3(6)   + 4(4)   = 18 + 16 =   34 (–1, –3):  z = 3(–1) + 4(–3) = –3 – 12 = –15

Then the maximum of z = 34 occurs at (6, 4), and the minimum of z = –15 occurs at (–1, –3).

2. Given the following constraints, maximize and minimize the value of z= –0.4x + 3.2y.

First I'll solve the fourth and fifth constraints for easier graphing:

The feasibility region looks like this:

From the graph, I can see which lines cross to form the corners, so I know which lines to pair up in order to verify the coordinates. I'll start at the "top" of the shaded area and work my way clockwise around the edges:

y = –x + 7y = x + 5

y = –x + 7x = 5

x = 5y = 0

–x + 7 = x + 52 = 2x1 = x

y = (1) + 5 = 6

y = –(5) + 7 = 2 [nothing to do]

corner at (1, 6) corner at (5, 2) corner at (5, 0)

y = 0y = –( 1/2 )x + 2

y = –( 1/2 )x + 2x = 0

x = 0y = x + 5

–( 1/2 )x + 2 = 02 = (1/2)x

4 = x

y = –( 1/2 )(0) + 2y = 0 + 2y = 2

y = (0) + 5 = 5

corner at (4, 0) corner at (0, 2) corner at (0, 5)

Now I'll plug each corner point into the optimization equation, z = –0.4x + 3.2y:

(1, 6):  z = –0.4(1) + 3.2(6) = –0.4 + 19.2 = 18.8 (5, 2):  z = –0.4(5) + 3.2(2) = –2.0 + 6.4   =   4.4 (5, 0):  z = –0.4(5) + 3.2(0) = –2.0 + 0.0   = –2.0 (4, 0):  z = –0.4(4) + 3.2(0) = –1.6 + 0.0   = –1.6 (0, 2):  z = –0.4(0) + 3.2(2) = –0.0 + 6.4   =   6.4 (0, 5):  z = –0.4(0) + 3.2(5) = –0.0 + 16.0 = 16.0

Then the maximum is 18.8 at (1, 6) and the minimum is –2 at (5, 0).

Given the inequalities, linear-programming exercise are pretty straightforward, if sometimes a bit long. The hard part is usually the word problems, where you have to figure out what the inequalities are. So I'll show how to set up some typical linear-programming word problems.

3. At a certain refinery, the refining process requires the production of at least two gallons of gasoline for each gallon of fuel oil. To meet the anticipated demands of winter, at least three million gallons of fuel oil a day will need to be produced. The demand for gasoline, on the other hand, is not more than 6.4 million gallons a day.

If gasoline is selling for $1.90 per gallon and fuel oil sells for $1.50/gal, how much of each should be produced in order to maximize revenue?

The question asks for the number of gallons which should be produced, so I should let my variables stand for "gallons produced".

x: gallons of gasoline producedy: gallons of fuel oil produced

Since this is a "real world" problem, I know that I can't have negative production levels, so the variables can't be negative. This gives me my first two constraints: namely, x > 0 and y > 0.

Since I have to have at least two gallons of gas for every gallon of oil, then x > 2y.

For graphing, of course, I'll use the more manageable form "y < ( 1/2 )x".

The winter demand says that y > 3,000,000; note that this constraint eliminates the need for the "y > 0" constraint. The gas demand says that x < 6,400,000.

I need to maximize revenue R, so the optimization equation is R = 1.9x + 1.5y. Then the model for this word problem is as follows:

R = 1.9x + 1.5y, subject to: x > 0 x < 6,400,000   Copyright © Elizabeth Stpe06-2011 All Rights Reserved

y > 3,000,000 y < ( 1/2 )x

Using a scale that counts by millions (so "y = 3" on the graph means "y is three million"), the above system graphs as follows:

Taking a closer look, I can see the feasibility region a little better:

When you test the corner points at (6.4m, 3.2m), (6.4m, 3m), and (6m, 3m), you should get a maximal solution of R = $16.96m at (x, y) = (6.4m, 3.2m).

4. A calculator company produces a scientific calculator and a graphing calculator. Long-term projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each day. Because of limitations on production capacity, no more than 200 scientific and 170graphing calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much be shipped each day.

If each scientific calculator sold results in a $2 loss, but each graphing calculator produces a $5 profit, how many of each type should be made daily to maximize net profits?

The question asks for the optimal number of calculators, so my variables will stand for that:

x: number of scientific calculators producedy: number of graphing calculators produced

Since they can't produce negative numbers of calculators, I have the two constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have that x > 100 andy > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum shipping requirement gives mex + y > 200; in other words, y > –x + 200. The revenue relation will be my optimization equation: R = –2x + 5y. So the entire system is:

R = –2x + 5y, subject to: 100 < x < 200 80 <  y < 170 y > –x + 200 

The feasibility region graphs as:   Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved

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When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80), and (100, 100), you should obtain the maximum value of R = 650 at (x, y) = (100, 170). That is, the solution is "100scientific calculators and 170 graphing calculators".

You need to buy some filing cabinets. You know that Cabinet X costs $10 per unit, requires six square feet of floor space, and holds eight cubic feet of files. Cabinet Y costs $20 per unit, requires eight square feet of floor space, and holds twelve cubic feet of files. You have been given $140 for this purchase, though you don't have to spend that much. The office has room for no more than 72 square feet of cabinets. How many of which model should you buy, in order to maximize storage volume?

The question ask for the number of cabinets I need to buy, so my variables will stand for that:

x: number of model X cabinets purchasedy: number of model Y cabinets purchased

Naturally, x > 0 and y > 0. I have to consider costs and floor space (the "footprint" of each unit), while maximizing the storage volume, so costs and floor space will be my constraints, while volume will be my optimization equation.

cost: 10x + 20y < 140, or y < –( 1/2 )x + 7 space: 6x + 8y < 72, or y < –( 3/4 )x + 9 volume: V = 8x + 12y

This system (along with the first two constraints) graphs as:

When you test the corner points at (8, 3), (0, 7), and (12, 0), you should obtain a maximal volume of100 cubic feet by buying eight of model X and three of model Y.