## How Do I Set Up A System Of Inequalities (Part 2)?

The most challenging part of these systems of inequalities is writing them out. The key is to write out the variables explicitly. This forces you to read the problem carefully as well as look for relationships between the different quantities in the problem.

Let’s look at another example where we wish to write and graph a system of inequalities corresponding to an application.

Problem A retiree wishes to invest in bonds and stocks. He has a total of 1.2 million dollars to invest. Bonds earn an average of 3% per year and stocks return an average of 8% per year. The retiree needs at least 60,000 dollars in returns each year to live on. Graph a system of inequalities that indicates all possible amounts that may invested in bonds and stocks. Label the corner points on the solution.

Solution Since we need to find all possible amounts invested in bonds and stocks, define the variables as

B: amount invested in bonds (in millions of dollars)

S: amount invested in stocks (in millions of dollars)

The retiree has a maximum of 1.2 million. Assuming that the retiree may not invest all of the money,

B + S < 1.2 million dollars

The return on each amount if found by multiplying the rates of return by the amounts. To earn at least 60,000 dollars,

0.03B + 0.08S > 0.06 million dollars

We can also assume that both amount cannot be negative so

B > 0     S > 0

Putting these inequalities together gives the system of inequalities,

B + S < 1.2

0.03B + 0.08S > 0.06

B > 0

S > 0

Graph the corresponding system of equalities.

Since (0, 0) is on the border of the region, let’s test the point (0.1, 0.1) in each inequality.

This corresponds to the solutions indicated below.

These solutions overlap in the triangular region below.

The corner points are located at the values labeled below.

Any point in the region or on its border satisfies all of the inequalities in the system. Any point along the blue border means all 1.2 million dollars will be invested. In the gray portion, only a portion is invested. However, since points inside of the region are above the red border, the retiree will earn more than 60,000 dollars in returns.

On the red border, the retiree will earn exactly 60,000 dollars, but will not invest the entire amount. At the corner point (0.72, 0.48), the investor will invest 720,000 dollars in bonds and 480,000 dollars in stocks. At this point the retiree will also have exactly 60,000 dollars in returns.

## How Do I Set Up A System Of Inequalities?

Section 4.1 focused on graphing a system of inequalities to find the solution. This involved graphing each inequality and combining the graphs to find what is in common between each of the inequalities.

Although this process is a bit tedious, it is not incredibly difficult. The Application Quiz for this section focuses on the more difficult part…where does the system come from in the first place. There are a number of examples in Section 4.1 to help guide you. Let’s work through another example to augment those applications.

Problem A small brewery supplies two distributors in the Southwest US. One distributor needs at least 30 kegs each month and the other needs at least 50 kegs each month. The brewery can produce at most 100 kegs each month. Graph a system of inequalities that indicates all possible amounts that may be sent to each distributor. Label the corner points on the solution.

Solution As with all application problems, we need to start by defining the variables. What are we being asked to find? The question indicates that we are interested in “all possible amounts that may be sent to each distributor”. So define

x1: number of kegs sent to the first distributor each month

x2: number of kegs sent to the second distributor each month

Since the first distributor needs at least 30 kegs, we write down

x1 > 30

The second distributor needs at least 50 kegs so we write

x2 > 50

In each case, “at least” means that the variables must be larger than or equal to the number so a > is used.

The brewery can produce no more than 100 kegs per month. This means that the amount sent to distributor 1 plus the amount sent to distributor 2 must be less than or equal to 100,

x1 + x2 < 100

We also know that the number of kegs must be nonnegative. However, we do not need nonnegativity constraints since the first two inequalities imply that both variables will not be negative.

Let’s write the three inequalities down.

x1 > 30

x2 > 50

x1 + x2 < 100

To graph the solution to this system of inequalities, let’s look at the corresponding equalities.

x1 = 30

x2 = 50

x1 + x2 = 100

Each of these equations corresponds to a line. Let’s graph those lines with x1 on the horizontal axis and x2 on the vertical axis. There is nothing special about labeling this way. We could have also reversed the labels and solved the system.

Now let’s test the point (x1, x2) = (0, 0) in each inequality.

Using this information, we can use arrows to indicate how the solutions for each individual inequality is shaded.

These shadings overlap in the triangular area in the center.

The solution to the system of inequalities is the shaded area. Using pairs of lines we can located the corner points.