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# Modeling with Linear Functions

Module by: First Last. E-mail the author

Summary: In this section, you will:

• Identify steps for modeling and solving.
• Build linear models from verbal descriptions.
• Build systems of linear models.

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Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending$400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

## Identifying Steps to Model and Solve Problems

When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
5. When needed, write a formula for the function.
6. Solve or evaluate the function using the formula.
7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
8. Clearly convey your result using appropriate units, and answer in full sentences when necessary.

## Building Linear Models

Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.

• Output: M, M, money remaining, in dollars
• Input: t, t, time, in weeks

So, the amount of money remaining depends on the number of weeks: M(t) M(t)

We can also identify the initial value and the rate of change.

• Initial Value: She saved $3,500, so$3,500 is the initial value for M. M.
• Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant, so we can start with the linear model M( t )=mt+b. M( t )=mt+b. Then we can substitute the intercept and slope provided.

To find the x- x- intercept, we set the output to zero, and solve for the input.

0=400t+3500 t= 3500 400 =8.75 0=400t+3500 t= 3500 400 =8.75

The x- x- intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved$3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x- x- intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0t8.75. 0t8.75.

In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

### Using a Given Intercept to Build a Model

Some real-world problems provide the y- y- intercept, which is the constant or initial value. Once the y- y- intercept is known, the x- x- intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay$250 per month until her balance is $0. The y- y- intercept is the initial amount of her debt, or$1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model. f(x)=mx+b =250x+1000 f(x)=mx+b =250x+1000 Now we can set the function equal to 0, and solve for x x to find the x- x- intercept. 0=250x+1000 1000=250x 4=x x=4 0=250x+1000 1000=250x 4=x x=4 The x- x- intercept is the number of months it takes her to reach a balance of$0. The x- x- intercept is 4 months, so it will take Hannah four months to pay off her loan.

### Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

#### How To:

Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

1. Identify the input and output values.
2. Convert the data to two coordinate pairs.
3. Find the slope.
4. Write the linear model.
5. Use the model to make a prediction by evaluating the function at a given x- x- value.
6. Use the model to identify an x- x- value that results in a given y- y- value.

#### Example 1

##### Problem 1
###### Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.

1. Predict the population in 2013.
2. Identify the year in which the population will reach 15,000.
###### Solution

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y- y- intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: t, t, years since 2004
• Output: P(t), P(t), the town’s population

To predict the population in 2013 (t=9), (t=9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

m= change in output change in input m= change in output change in input

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t=0, t=0, giving the point ( 0,6200 ). ( 0,6200 ). Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t=5, t=5, giving the point ( 5,8100 ). ( 5,8100 ).

The two coordinate pairs are ( 0,6200 ) ( 0,6200 ) and ( 5,8100 ). ( 5,8100 ). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

m= 81006200 50    = 1900 5    =380 people per year m= 81006200 50    = 1900 5    =380 people per year

We already know the y-intercept of the line, so we can immediately write the equation:

P(t)=380t+6200 P(t)=380t+6200

To predict the population in 2013, we evaluate our function at t=9. t=9.

P(9)=380(9)+6,200       =9,620 P(9)=380(9)+6,200       =9,620

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set P(t)=15000 P(t)=15000 and solve for t. t.

15000=380t+6200   8800=380t          t23.158 15000=380t+6200   8800=380t          t23.158

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

#### Try It:

##### Exercise 1

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs$0.25 to produce each doughnut.

1. Write a linear model to represent the cost C C of the company as a function of x, x, the number of doughnuts produced.
2. Find and interpret the y-intercept.

#### Exercise 47

If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

### Real-World Applications

#### Exercise 48

In 2004, a school population was 1001. By 2008 the population had grown to 1697. Assume the population is changing linearly.

1. How much did the population grow between the year 2004 and 2008?
2. How long did it take the population to grow from 1001 students to 1697 students?
3. What is the average population growth per year?
4. What was the population in the year 2000?
5. Find an equation for the population, P, P, of the school t years after 2000.
6. Using your equation, predict the population of the school in 2011.
##### Solution

a. 696 people; b. 4 years; c. 174 people per year; d. 305 people; e. P(t)=305+174t P(t)=305+174t ; f. 2219 people

#### Exercise 49

In 2003, a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly.

1. How much did the population grow between the year 2003 and 2007?
2. How long did it take the population to grow from 1431 people to 2134 people?
3. What is the average population growth per year?
4. What was the population in the year 2000?
5. Find an equation for the population, PP of the town tt years after 2000.
6. Using your equation, predict the population of the town in 2014.

#### Exercise 50

A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be$118.

1. Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used.
2. Interpret the slope and y-intercept of the equation.
3. Use your equation to find the total monthly cost if 687 minutes are used.

a. C( x )=0.15x+10 C( x )=0.15x+10 ; b. The flat monthly fee is $10 and there is an additional$0.15 fee for each additional minute used; c. $113.05 #### Exercise 51 A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of$10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be$17.80.

1. Find a linear equation for the monthly cost of the data plan as a function of xx, the number of MB used.
2. Interpret the slope and y-intercept of the equation.
3. Use your equation to find the total monthly cost if 250 MB are used.

#### Exercise 52

In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly.

1. Find a formula for the moose population, P since 1990.
2. What does your model predict the moose population to be in 2003?
##### Solution

a. P( t )=190t+4360 P( t )=190t+4360 ; b. 6640 moose

#### Exercise 53

In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990.

1. Find a formula for the owl population, P.P. Let the input be years since 2003.
2. What does your model predict the owl population to be in 2012?

#### Exercise 54

The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year.

1. Give a linear equation for the remaining federal helium reserves, R,R, in terms of t,t, the number of years since 2010.
2. In 2015, what will the helium reserves be?
3. If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?
##### Solution

a. R( t )=162.1t R( t )=162.1t ; b. 5.5 billion cubic feet; c. During the year 2017

#### Exercise 55

Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year:

1. Give a linear equation for the remaining oil reserves, R,R, in terms of t,t, the number of years since now.
2. Seven years from now, what will the oil reserves be?
3. If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?

#### Exercise 58

When hired at a new job selling jewelry, you are given two pay options:

• Option A: Base salary of $17,000 a year with a commission of 12% of your sales • Option B: Base salary of$20,000 a year with a commission of 5% of your sales

How much jewelry would you need to sell for option A to produce a larger income?

#### Exercise 61

When hired at a new job selling electronics, you are given two pay options:

• Option A: Base salary of $10,000 a year with a commission of 9% of your sales • Option B: Base salary of$20,000 a year with a commission of 4% of your sales

How much electronics would you need to sell for option A to produce a larger income?

## Footnotes

1. Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/

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