1.4 Modeling with Linear Functions

Learning Objectives

Construct and solve applications using linear functions.

To use a linear function to describe a real-world application, we must first determine the known quantities and define the unknown quantities as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. For example, think of the rental price for a car. Say the company charges [latex]$0.10[/latex] per mile in addition to a flat rate. In this case, to find the total cost to rent the car the known cost of [latex]$0.10[/latex] per mile will be multiplied by an unknown quantity; the number of miles driven. If we use [latex]x[/latex] to represent the unknown number of miles driven, then [latex]0.10x[/latex] represents the variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. If the car rental company charges [latex]$0.10[/latex] per mile plus a daily fee of [latex]$50[/latex], then the equation that describes the daily car rental cost, [latex]C[/latex], is [latex]C=0.10x+50[/latex].

Creating a Model

At the beginning of section 1.1, we discussed that climate change is one of the most serious threats to our planet today and how the increasing levels of CO₂ in the earth’s atmosphere parallels the increase in global temperatures. In 2016, the global temperature was $$1.1$$ degrees C ($$2$$ degrees F) warmer than pre-industrial levels. The Paris Agreement was made and aims to keep the global temperature increase well below $$2$$ degrees C, and hopefully limit that increase to $$1.5$$ degrees C. (https://e360.yale.edu/features/how-the-world-passed-a-carbon-threshold-400ppm-and-why-it-matters) Although 196 countries signed this agreement, global warming is stilling increasing. From 2015 – 2019 the global temperature has risen $$0.2$$ degrees C. If this trend keeps rising like it has, our planet will be in serious danger.

We can use data from the follow graph to make a linear equation that models the increase of CO₂ levels and use that model to predict the CO₂ levels in the future, if the CO₂ levels keep increasing at a similar rate. Data on CO₂ levels have been collected since 1958 and is presented graphically below (red graph). We will be focusing on data starting with January 2005 up to present time. When we look at that portion of the graph the trend appears to follow an overall increasing linear model.

This link https://gml.noaa.gov/ccgg/trends/graph.html will take you to an interactive graph that you can use to fill out the following table.

Date of Monthly Averages	CO₂ levels (in ppm)	Ordered Pair $$(x,y)$$
Jan 2005	378.63	(2005, 378.63)
Jan 2008
Jan 2011
Jan 2014
Jan 2017
Jan 2020
Jan 2023

Let’s find the average rate of change from Jan 2005 to Jan 2014, using the slope formula. Recall that the slope [latex]m[/latex], or rate of change of a linear function between two points [latex]\left({x}_{2,\text{ }}{y}_{2}\right)[/latex] and [latex]\left({x}_{1},\text{ }{y}_{1}\right)[/latex] is: [latex]\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex].

$$\dfrac{Change\ in\ CO_{2}\ levels}{Change\ in\ years} = \dfrac{398.04-378.63}{2014-2005} = \dfrac{19.41}{9} = 2.16$$

The average CO₂ increase per year is $$2.16$$ ppm.

We can use this information to write a linear equation that models the increase of atmospheric CO₂ since Jan 2005. Let $$x$$ represent the number of years since 2005 ($$x=0$$ for Jan 2005) and let $$y$$ represent the amount of atmospheric CO₂ in that given year, measured in parts per million (ppm).

We will write this equation in $$y=mx+b$$ form. As discussed earlier in this module, $$m$$ is the slope and $$b$$ is the $$y$$-intercept. When we are writing a linear model, $$m$$ is the average rate of change and $$b$$ is the start amount (or initial value). Above, we found the average rate of change of CO₂ to be $$2.16$$ ppm. The start amount is the amount of atmospheric CO₂ in Jan 2005, which is $$378.63$$ ppm.

Now substituting these values in place of $$m$$ and $$b$$ we get $$y=2.16x+378.63$$.

We can now use this linear model to make some predictions (assuming that the average rate of change stays the same).

Predict the amount of atmospheric CO₂ (in ppm) in January 2030 and January 2045

In Jan 2030: In order to use the above linear model we need to determine $$x$$. Remember that $$x=0$$ for Jan 2005. To find the $$x$$ we will find the difference in the years, $$x=2030-2005=25$$. Now substitute $$\color{Green}{25}$$ in place of $$\color{Green}{x}$$ and solve for $$y$$.

$$\begin{align} y &= 2.16\color{Green}{x}\color{black}{+378.63}\\ y &= 2.16(\color{Green}{25}\color{black}{)+378.63}\\ y &= 54+378.63\\ y &= 432.63 \end{align}$$

So in 2030, the CO₂ levels will have risen to approximately $$432.63$$ ppm.

In Jan 2045: To find the $$x$$ we will find the difference of the years, $$x=2045-2005=\color{Green}{40}$$.

$$\begin{align} y &= 2.16x+378.63\\ y &= 2.16(\color{Green}{40}\color{black}{)+378.63}\\ y &= 86.4+378.63\\ y &= 465.03 \end{align}$$

So in 2045, the CO₂ levels will have risen to approximately $$465.03$$ ppm.

When will the atmospheric CO₂ levels reach $$500$$ ppm, according to the model? (Round to nearest year)

Replace $$\color{Green}{y}$$ with $$\color{Green}{500}$$ and solve for x.

$$\begin{align} y &= 2.16x+378.63\\ \color{Green}{500} &= 2.16x+378.63\\ 500-378.63 &= 2.16x && \color{blue}{\textsf{subtract}}\\ 121.37 &= 2.16x && \color{blue}{\textsf{divide}}\\ 56.189… &= x \end{align}$$

So, the CO₂ levels will reach $$500$$ ppm approximately 56 years after 2005. The CO₂ levels will be over $$500$$ ppm in the year 2061.

Initial Value and Rate of Change

Problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know or can figure out the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

How To: Given a linear function [latex]f[/latex] and the initial value and rate of change, evaluate [latex]f(c)[/latex]

Determine the start amount (initial value) and the rate of change (slope).
Substitute the values into [latex]f\left(x\right)=mx+b[/latex].
Evaluate the function at [latex]x=c[/latex].

Initial value is a term that is typically used in applications of functions sometimes called the start amount. It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the $$y$$-intercept. Here are some characteristics of the initial value:

The point [latex](0,y)[/latex] is often the initial value of a linear function
The $$y$$-value of the initial value comes from $$b$$ in slope-intercept form of a linear function, [latex]f\left(x\right)=mx+b[/latex]
The initial value can be found by solving for $$b$$ or substituting $$0$$ in for $$x$$ in a linear function.

Example

Marcus currently has [latex]200[/latex] songs in his music collection. Every month he adds [latex]15[/latex] new songs. Write a formula for the number of songs, $$N$$, in his collection as a function of time, $$t$$, the number of months.

1) How many songs will he own in a year?

2) How many months will it take to grow his collection to 500 songs?

Show Solution

In the example above, notice that $$N$$ is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Comparisons with Linear Functions

Example

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of [latex]$34[/latex] plus [latex]$0.05[/latex] per minute of talk-time. Company B charges a monthly service fee of [latex]$40[/latex] plus [latex]$0.04[/latex] per minute of talk-time.

Write a linear equation that models the packages offered by both companies.
If the average number of minutes used each month is [latex]1,160[/latex], which company offers the better plan?
If the average number of minutes used each month is [latex]420[/latex], which company offers the better plan?

Show Solution

The following video shows another example of writing two equations that will allow you to compare two different cell phone plans.

Writing a Linear Function Given Two Data Points

The next example demonstrating writing a linear function given two data points. In this case, Ilya’s weekly income depends on the number of insurance policies he sells. We are given his income for two different weeks and the number of policies sold. We first find the rate of change and then solve for the initial value.

Example

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, $$I$$, depends on the number of new policies, $$n$$, he sells during the week. Last week, he sold [latex]3[/latex] new policies and earned [latex]$760[/latex] for the week. The week before, he sold [latex]5[/latex] new policies and earned [latex]$920[/latex].

1) Find an equation for $$I(n)$$ and interpret the meaning of the components of the equation.

2) What is Ilya’s weekly income if he sells $$12$$ policies?

3) How many policies does he need to sell to make a weekly income of $2040?

Show Solution

1) The given information gives us two input-output pairs (number of policies sold, amount of money earned: [latex](3, 760)[/latex] and [latex](5, 920)[/latex]. We start by finding the rate of change.

Recall that the slope [latex]m[/latex], or rate of change of a linear function between two points [latex]\left({x}_{1,\text{ }}{y}_{1}\right)[/latex] and [latex]\left({x}_{2},\text{ }{y}_{2}\right)[/latex] is: [latex]\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex].

[latex]\begin{align}m&=\dfrac{920 - 760}{5 - 3}\\&=\dfrac{$160}{\text{2 policies}}\\&=$80\text{ per policy}\end{align}[/latex]

Keeping track of units can help us interpret this quantity. Income increased by [latex]$160[/latex] when the number of policies increased by [latex]2[/latex], so the rate of change is [latex]$80[/latex] per policy. Therefore, Ilya earns a commission of [latex]$80[/latex] for each policy sold during the week.

We can then solve for the initial value.

[latex]\begin{array}\text{ }I\left(n\right)=80n+b\hfill & \hfill \\ \text{ }760=80\left(3\right)+b\hfill & \text{when }n=3, I\left(3\right)=760\hfill \\ 760 - 80\left(3\right)=b\hfill & \hfill \\ \text{ }520=b\hfill & \hfill \end{array}[/latex]

The value of $$b$$ is the starting value for the function and represents Ilya’s income when [latex]n = 0[/latex], or when no new policies are sold. We can interpret this as Ilya’s base salary for the week which does not depend upon the number of policies sold.

We can now write the final equation.

[latex]I\left(n\right)=80n+520[/latex]

We can interpret the equation that Ilya’s base salary is [latex]$520[/latex] per week and he earns an additional [latex]$80[/latex] commission for each policy sold.

2) To find Ilya’s weekly income if he sells $$12$$ policies, let’s use the function $$I(n)=80n+520$$. Start by replacing $$n$$ with $$12$$ in the function, then solve for $$I(n)$$.

$$\begin{align}
I(\color{Green}{n}) &= 80\color{Green}{n} + 520 \\
I(\color{Green}{12}) &= 80(\color{Green}{12}) + 520 \\
&= 960 + 520 \\
&= 1480
\end{align}$$

Ilya will make [latex]$1480[/latex] if he sells $$12$$ policies.

3) If Ilya wants to make [latex]$2040[/latex] a week, we need to find how many policies he would need to sell. Replace $$I(n)$$ with $$2040$$ and then solve for $$n$$.

$$\begin{align}
\color{Green}{I(n)} &= 80n + 520 \\
\color{Green}{2040} &= 80n + 520 \\
2040-520 &= 80n \\
1520 &= 80n \\ 19 &= n
\end{align}$$

Ilya will need to sell $$19$$ policies if he wants to make [latex]$2040[/latex].

In the following video example, we show how to identify the initial value, slope, and equation for a linear function.

Modeling Monthly Cost

This example of modeling real-world scenarios using linear functions involves the monthly cost to run a company given monthly fixed costs and production costs per item manufactured.

Example

Suppose Sofia starts a company in which she incurs a fixed cost of [latex]$1,250[/latex] per month for the overhead which includes office rent. Her production costs are [latex]$37.50[/latex] per item. Write a linear function $$C$$ where $$C(x)$$ is the cost for $$x$$ items produced in a given month. What will be the monthly cost if Sofia produces $$100$$ items in a month?

Show Solution

A final video example shows how to write a linear function that represents how many miles you can travel in a rental car given a fixed amount of money.

Module 1: Graphs, Functions, and Linear Equations

Learning Objectives

Creating a Model

Initial Value and Rate of Change

How To: Given a linear function [latex]f[/latex] and the initial value and rate of change, evaluate [latex]f(c)[/latex]

Example

Comparisons with Linear Functions

Example

Writing a Linear Function Given Two Data Points

Example

Modeling Monthly Cost

Example