- Graph linear inequalities in two variables.
- Solve systems of linear inequalities.
- Solve optimization problems.
In the last chapter you learned how to graph a linear inequality in two variables. To do that, you graphed the equation of the straight line on the coordinate plane. The line was solid for ≤ or ≥ signs (where the equals sign is included), and the line was dashed for < or > signs (where the equals sign is not included). Then you shaded above the line (if the inequality began with y> or y≥) or below the line (if it began with y< or y≤).
In this section, we’ll see how to graph two or more linear inequalities on the same coordinate plane. The inequalities are graphed separately on the same graph, and the solution for the system is the common shaded region between all the inequalities in the system. One linear inequality in two variables divides the plane into two half-planes. A system of two or more linear inequalities can divide the plane into more complex shapes.
Let’s start by solving a system of two inequalities.
Graph a System of Two Linear Inequalities
Solve the following system:
Solving systems of linear inequalities means graphing and finding the intersections. So we graph each inequality, and then find the intersection regions of the solution.
First, let’s rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell which region of the coordinate plane to shade.) Our system becomes
Notice that the inequality sign in the second equation changed because we divided by a negative number!
For this first example, we’ll graph each inequality separately and then combine the results.
Here’s the graph of the first inequality:
The line is solid because the equals sign is included in the inequality. Since the inequality is less than or equal to, we shade below the line.
And here’s the graph of the second inequality:
The line is solid again because the equals sign is included in the inequality. We now shade above the line because y is greater than or equal to.
When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they overlap is the area where both inequalities are true. Thus that area (shown below in purple) is the solution of the system.
The kind of solution displayed in this example is called unbounded, because it continues forever in at least one direction (in this case, forever upward and to the left).
There are also situations where a system of inequalities has no solution. For example, let’s solve this system.
We start by graphing the first line. The line will be solid because the equals sign is included in the inequality. We must shade downwards because y is less than.
Next we graph the second line on the same coordinate axis. This line will be dashed because the equals sign is not included in the inequality. We must shade upward because y is greater than.
It doesn’t look like the two shaded regions overlap at all. The two lines have the same slope, so we know they are parallel; that means that the regions indeed won’t ever overlap since the lines won’t ever cross. So this system of inequalities has no solution.
But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what happens if we swap the directions of the inequality signs in the system we just graphed?
To graph the system
we draw the same lines we drew for the previous system, but we shade upward for the first inequality and downward for the second inequality. Here is the result:
You can see that this time the shaded regions overlap. The area between the two lines is the solution to the system.
Graph a System of More Than Two Linear Inequalities
When we solve a system of just two linear inequalities, the solution is always an unbounded region—one that continues infinitely in at least one direction. But if we put together a system of more than two inequalities, sometimes we can get a solution that is bounded—a finite region with three or more sides.
Let’s look at a simple example.
Find the solution to the following system of inequalities.
Let’s start by writing our inequalities in slope-intercept form.
Now we can graph each line and shade appropriately. First we graph y>3x−4 :
Next we graph y<−94x+2 :
Finally we graph x≥0 and y≥0, and we’re left with the region below; this is where all four inequalities overlap.
The solution is bounded because there are lines on all sides of the solution region. In other words, the solution region is a bounded geometric figure, in this case a triangle.
Notice, too, that only three of the lines we graphed actually form the boundaries of the region. Sometimes when we graph multiple inequalities, it turns out that some of them don’t affect the overall solution; in this case, the solution would be the same even if we’d left out the inequality y>3x−4. That’s because the solution region of the system formed by the other three inequalities is completely contained within the solution region of that fourth inequality; in other words, any solution to the other three inequalities is automatically a solution to that one too, so adding that inequality doesn’t narrow down the solution set at all.
But that wasn’t obvious until we actually drew the graph!
Solve Real-World Problems Using Systems of Linear Inequalities
A lot of interesting real-world problems can be solved with systems of linear inequalities.
For example, you go to your favorite restaurant and you want to be served by your best friend who happens to work there. However, your friend only waits tables in a certain region of the restaurant. The restaurant is also known for its great views, so you want to sit in a certain area of the restaurant that offers a good view. Solving a system of linear inequalities will allow you to find the area in the restaurant where you can sit to get the best view and be served by your friend.
Often, systems of linear inequalities deal with problems where you are trying to find the best possible situation given a set of constraints. Most of these application problems fall in a category called linear programming problems.
Linear programming is the process of taking various linear inequalities relating to some situation, and finding the best possible value under those conditions. A typical example would be taking the limitations of materials and labor at a factory, then determining the best production levels for maximal profits under those conditions. These kinds of problems 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 this section, we’ll only work with the simple two-variable linear case.
The general process is to:
- Graph the inequalities (called constraints) to form a bounded area on the coordinate plane (called the feasibility region).
- Figure out the coordinates of the corners (or vertices) of this feasibility region by solving the system of equations that applies to each of the intersection points.
- Test these corner points in the formula (called the optimization equation) for which you're trying to find the maximum or minimum value.
If z=2x+5y, find the maximum and minimum values of z given these constraints:
First, we need to find the solution to this system of linear inequalities by graphing and shading appropriately. To graph the inequalities, we rewrite them in slope-intercept form:
These three linear inequalities are called the constraints, and here is their graph:
The shaded region in the graph is called the feasibility region. All possible solutions to the system occur in that region; now we must try to find the maximum and minimum values of the variable z within that region. In other words, which values of x and y within the feasibility region will give us the greatest and smallest overall values for the expression 2x+5y?
Fortunately, we don’t have to test every point in the region to find that out. It just so happens that the minimum or maximum value of the optimization equation in a linear system like this will always be found at one of the vertices (the corners) of the feasibility region; we just have to figure out which vertices. So for each vertex—each point where two of the lines on the graph cross—we need to solve the system of just those two equations, and then find the value of z at that point.
The first system consists of the equations y=2x−12 and y=−43x. We can solve this system by substitution:
−43xy=2x−12⇒−4x=6x−36⇒−10x=−36⇒x=3.6=2x−12⇒y=2(3.6)−12⇒y=−4.8 The lines intersect at the point (3.6, -4.8).
The second system consists of the equations y=2x−12 and y=x−6. Solving this system by substitution:
x−6y=2x−12⇒6=x⇒x=6=x−6⇒y=6−6⇒y=6 The lines intersect at the point (6, 6).
The third system consists of the equations y=−43x and y=x−6. Solving this system by substitution:
x−6y=−43x⇒3x−18=−4x⇒7x=18⇒x=2.57=x−6⇒y=2.57−6⇒y=−3.43 The lines intersect at the point (2.57, -3.43).
So now we have three different points that might give us the maximum and minimum values for z. To find out which ones actually do give the maximum and minimum values, we can plug the points into the optimization equation z=2x+5y.
When we plug in (3.6, -4.8), we get z=2(3.6)+5(−4.8)=−16.8.
When we plug in (6, 0), we get z=2(6)+5(0)=12.
When we plug in (2.57, -3.43), we get z=2(2.57)+5(−3.43)=−12.01.
So we can see that the point (6, 0) gives us the maximum possible value for z and the point (3.6, –4.8) gives us the minimum value.
In the previous example, we learned how to apply the method of linear programming in the abstract. In the next example, we’ll look at a real-life application.
You have $10,000 to invest, and three different funds to choose from. The municipal bond fund has a 5% return, the local bank's CDs have a 7% return, and a high-risk account has an expected 10% return. To minimize risk, you decide not to invest any more than $1,000 in the high-risk account. For tax reasons, you need to invest at least three times as much in the municipal bonds as in the bank CDs. What’s the best way to distribute your money given these constraints?
Let’s define our variables:
x is the amount of money invested in the municipal bond at 5% return
y is the amount of money invested in the bank’s CD at 7% return
10000−x−y is the amount of money invested in the high-risk account at 10% return
z is the total interest returned from all the investments, so z=.05x+.07y+.1(10000−x−y) or z=1000−0.05x−0.03y. This is the amount that we are trying to maximize. Our goal is to find the values of x and y that maximizes the value of z.
Now, let’s write inequalities for the constraints:
You decide not to invest more than $1000 in the high-risk account—that means:
You need to invest at least three times as much in the municipal bonds as in the bank CDs—that means:
3y≤x Also, you can’t invest less than zero dollars in each account, so:
To summarize, we must maximize the expression z=1000−.05x−.03y using the constraints:
10000−x−y≤10003y≤xx≥0y≥010000−x−y≥0Or in slope-intercept form:y≥9000−xy≤x3x≥0y≥0y≤10000−x
Step 1: Find the solution region to the set of inequalities by graphing each line and shading appropriately. The following figure shows the overlapping region:
The purple region is the feasibility region where all the possible solutions can occur.
Step 2: Next we need to find the corner points of the feasibility region. Notice that there are four corners. To find their coordinates, we must pair up the relevant equations and solve each resulting system.
Substitute the first equation into the second equation:
The intersection point is (7500, 2500).
Substitute the first equation into the second equation:
The intersection point is (6750, 2250).
The intersection point is (10000, 0).
The intersection point is (9000, 0).
Step 3: In order to find the maximum value for z, we need to plug all the intersection points into the equation for z and find which one yields the largest number.
(7500, 2500): z=1000−0.05(7500)−0.03(2500)=550
(6750, 2250): z=1000−0.05(6750)−0.03(2250)=595
(10000, 0): z=1000−0.05(10000)−0.03(0)=500
(9000, 0): z=1000−0.05(9000)−0.03(0)=550
The maximum return on the investment of $595 occurs at the point (6750, 2250). This means that:
$6,750 is invested in the municipal bonds.
$2,250 is invested in the bank CDs.
$1,000 is invested in the high-risk account.
Graphing calculators can be very useful for problems that involve this many inequalities. The video at http://www.youtube.com/watch?v=__wAxkYmhvY shows a real-world linear programming problem worked through in detail on a graphing calculator, although the methods used there can also be used for pencil-and paper solving.
- Consider the system y<3x−5y>3x−5. Is it consistent or inconsistent? Why?
- Consider the system y≤2x+3y≥2x+3. Is it consistent or inconsistent? Why?
- Consider the system y≤−x+1y>−x+1. Is it consistent or inconsistent? Why?
- In example 3 in this lesson, we solved a system of four inequalities and saw that one of the inequalities, y>3x−4, didn’t affect the solution set of the system.
- What would happen if we changed that inequality to y<3x−4?
- What’s another inequality that we could add to the original system without changing it? Show how by sketching a graph of that inequality along with the rest of the system.
- What’s another inequality that we could add to the original system to make it inconsistent? Show how by sketching a graph of that inequality along with the rest of the system.
- Recall the compound inequalities in one variable that we worked with back in chapter 6. Compound inequalities with “and” are simply systems like the ones we are working with here, except with one variable instead of two.
- Graph the inequality x>3 in two dimensions. What’s another inequality that could be combined with it to make an inconsistent system?
- Graph the inequality x≤4 on a number line. What two-dimensional system would have a graph that looks just like this one?
Find the solution region of the following systems of inequalities.
Solve the following linear programming problems.
- Given the following constraints, find the maximum and minimum values of z=−x+5y: x+3y≤0x−y≥03x−7y≤16
- Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice elves draw a wage of five candy canes per hour worked, but can only make four trucks an hour. Senior elves can make six trucks an hour and are paid eight candy canes per hour. There’s only room for nine elves in the truck shop, and due to a candy-makers’ strike, Santa Claus can only pay out 480 candy canes for the whole 8-hour shift.
- How many senior elves and how many apprentice elves should work this shift to maximize the number of trucks that get made?
- How many trucks will be made?
- Just before the shift begins, the apprentice elves demand a wage increase; they insist on being paid seven candy canes an hour. Now how many apprentice elves and how many senior elves should Santa assign to this shift?
- How many trucks will now get made, and how many candy canes will Santa have left over?
- In Adrian’s Furniture Shop, Adrian assembles both bookcases and TV cabinets. Each type of furniture takes her about the same time to assemble. She figures she has time to make at most 18 pieces of furniture by this Saturday. The materials for each bookcase cost her $20 and the materials for each TV stand costs her $45. She has $600 to spend on materials. Adrian makes a profit of $60 on each bookcase and a profit of $100 on each TV stand.
- Set up a system of inequalities. What x− and y−values do you get for the point where Adrian’s profit is maximized? Does this solution make sense in the real world?
- What two possible real-world x−values and what two possible real-world y−values would be closest to the values in that solution?
- With two choices each for x and y, there are four possible combinations of x− and y−values. Of those four combinations, which ones actually fall within the feasibility region of the problem?
- Which one of those feasible combinations seems like it would generate the most profit? Test out each one to confirm your guess. How much profit will Adrian make with that combination?
- Based on Adrian’s previous sales figures, she doesn’t think she can sell more than 8 TV stands. Now how many of each piece of furniture should she make, and what will her profit be?
- Suppose Adrian is confident she can sell all the furniture she can make, but she doesn’t have room to display more than 7 bookcases in her shop. Now how many of each piece of furniture should she make, and what will her profit be?
- Here’s a “linear programming” problem on a line instead of a plane: Given the constraints x≤5 and x≥−2, maximize the value of y where y=x+3.
Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9617.