Modeling with Linear Functions

A panoramic view of Seattle, featuring the iconic Space Needle rising prominently in the foreground, with the sparkling waters of Puget Sound and the majestic, snow-capped Olympic Mountains stretching across the horizon under a clear blue sky. — Figure 1 (credit: EEK Photography/Flickr)

Elan is a college student who plans to spend a summer in Seattle. Elan has saved $3,500 for the trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent the situation? What would be the x-intercept, and what can Elan 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:

Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
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.
Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
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.
When needed, write a formula for the function.
Solve or evaluate the function using the formula.
Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
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 Elan's situation, there are two changing quantities: time and money. The amount of money they have remaining while on vacation depends on how long they stay. We can use this information to define our variables, including units.

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

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

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

Initial Value: They saved $3,500, so $3,500 is the initial value for $M .$
Rate of Change: They anticipate 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 they are spending money each week.

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

A diagram illustrates the substitution of m = -400 and b = 3500 into the linear equation M(t) = mt + b, resulting in the specific linear function M(t) = -400t + 3500.

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

\begin{array}{l} 0 = - 400 t + 3500 \\ t = \frac{3500}{400} \\ = 8.75 \end{array}

The $x -$ intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Elan 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 Elan saved $3,500, but the scenario discussed poses the question once they saved $3,500 because this is when the trip and subsequent spending starts. It is also likely that this model is not valid after the $x -$ intercept, unless Elan will use a credit card and go into debt. The domain represents the set of input values, so the reasonable domain for this function is $0 \leq t \leq 8.75.$

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 $y -$ intercept, which is the constant or initial value. Once the $y -$ intercept is known, the $x -$ 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 $y -$ 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.

\begin{array}{l} f (x) = m x + b \\ = - 250 x + 1000 \end{array}

Now we can set the function equal to 0, and solve for $x$ to find the $x -$ intercept.

\begin{array}{l} 0 = - 250 x + 1000 \\ 1000 = 250 x \\ 4 = x \\ x = 4 \end{array}

The $x -$ intercept is the number of months it takes her to reach a balance of $0. The $x$ -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.

Example 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.

ⓐ Predict the population in 2013.
ⓑ 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 $y -$ 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: $t,$ years since 2004
Output: $P (t),$ the town’s population

To predict the population in 2013 $(t = 9),$ 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 = \frac{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 $t = 0,$ giving the point $(0, 6200) .$ 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 $t = 5,$ giving the point $(5, 8100) .$

The two coordinate pairs are $(0, 6200)$ and $(5, 8100) .$ 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.

\begin{array}{l} \begin{array}{l} m = \frac{8100 - 6200}{5 - 0} \end{array} \\ = \frac{1900}{5} \\ = 380 people per year \end{array}

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

P (t) = 380 t + 6200

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

\begin{array}{l} P (9) = 380 (9) + 6, 200 \\ = 9, 620 \end{array}

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 $P (t) = 15000$ and solve for $t .$

\begin{array}{l} 15000 = 380 t + 6200 \\ 8800 = 380 t \\ t \approx 23.158 \end{array}

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

Using a Diagram to Model a Problem

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Example 2

Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

Solution

In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:

“How long will it take them to be 2 miles apart?”

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

Input: $t,$ time in hours.
Output: $A (t),$ distance in miles, and $E (t),$ distance in miles

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2.

Anna walks east at 4 mph, and Emanuel walks south at 3 mph. The diagram illustrates their perpendicular paths and the diagonal distance between them, forming a right triangle. — Figure 2

Initial Value: They both start at the same intersection so when $t = 0,$ the distance traveled by each person should also be 0. Thus the initial value for each is 0.

Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for $A$ is 4 and the slope for $E$ is 3.

Using those values, we can write formulas for the distance each person has walked.

\begin{array}{l} A (t) = 4 t \\ E (t) = 3 t \end{array}

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, $A,$ which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, $E,$ to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, $D,$ to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3.

Recall that we need to know how long it takes for $D,$ the distance between them, to equal 2 miles. Notice that for any given input $t,$ the outputs $A (t), E (t),$ and $D (t)$ represent distances.

Diagram showing a man looking at a woman, with horizontal distance "A", vertical distance "E", and direct line of sight "D" forming a right triangle. — Figure 3

Figure 2 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

\begin{array}{l} D {(t)}^{2} = A {(t)}^{2} + E {(t)}^{2} \\ = {(4 t)}^{2} + {(3 t)}^{2} \\ = 16 t^{2} + 9 t^{2} \\ = 25 t^{2} \\ D (t) = \pm \sqrt{25 t^{2}} & Solve for D (t) using the square root \\ = \pm 5 | t | \end{array}

In this scenario we are considering only positive values of $t,$ so our distance $D (t)$ will always be positive. We can simplify this answer to $D (t) = 5 t .$ This means that the distance between Anna and Emanuel is also a linear function. Because $D$ is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output $D (t) = 2$ and solve for $t .$

\begin{array}{l} D (t) = 2 \\ 5 t = 2 \\ t = \frac{2}{5} = 0.4 \end{array}

They will fall out of radio contact in 0.4 hours, or 24 minutes.

Example 3

Using a Diagram to Model Distance between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

Solution

It might help here to draw a picture of the situation. See Figure 4. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates $(30, 10),$ and Eastborough at $(20, 0) .$

A diagram depicting a coordinate plane with the origin (0,0) labeled as Westborough. Eastborough is located at (20,0) on the x-axis, 20 miles from Westborough. Agritown is located at (30,10). A blue line segment connects Westborough to Agritown. Another blue line segment extends perpendicularly from Eastborough to the line segment connecting Westborough and Agritown, indicated by a right angle symbol. — Figure 4

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:

m = \frac{10 - 0}{30 - 0} = \frac{1}{3}

The equation of the road from Westborough to Agritown would be

W (x) = \frac{1}{3} x

From this, we can determine the perpendicular road to Eastborough will have slope $m = - 3.$ Because the town of Eastborough is at the point (20, 0), we can find the equation:

\begin{array}{l} E (x) = - 3 x + b \\ 0 = - 3 (20) + b & Substitute in (20, 0) \\ b = 60 \\ E (x) = - 3 x + 60 \end{array}

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

\begin{array}{l} \frac{1}{3} x = - 3 x + 60 \\ \frac{10}{3} x = 60 \\ 10 x = 180 \\ x = 18 & Substituting this back into W (x) \\ y = W (18) \\ = \frac{1}{3} (18) \\ = 6 \end{array}

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

\begin{array}{l} distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \\ = \sqrt{{(18 - 0)}^{2} + {(6 - 0)}^{2}} \\ \approx 18.974 miles \end{array}

Analysis

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

Building Systems of Linear Models

Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5.

Graphs illustrating the three possible outcomes for a system of two linear equations: (a) intersecting lines (one solution), (b) coincident lines (infinitely many solutions), and (c) parallel lines (no solutions). — Figure 5

Example 4

Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mileRates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/. When will Keep on Trucking, Inc. be the better choice for Jamal?

Solution

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

Table 1 Three rows and three columns. In the first column, are the years 1950 and 2000. In the second columns are the house values for Indiana, which are 37700 for 1950 and 94300 for 2000. In the third columns are the house values for Alabama, which are 27100 for 1950 and 85100 for 2000.
Input	$d,$ distance driven in miles
Outputs	$K (d) :$ cost, in dollars, for renting from Keep on Trucking $M (d)$ cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: $K (0) = 20$ and $M (0) = 16$
Rate of Change	$K (d) = $ 0. 59$ /mile and $P (d) = $ 0. 63$ /mile

A linear function is of the form $f (x) = m x + b .$ Using the rates of change and initial charges, we can write the equations

\begin{matrix} K (d) = 0.59 d + 20 \\ M (d) = 0.63 d + 16 \end{matrix}

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when $K (d) < M (d) .$ The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the $K (d)$ function is smaller.

These graphs are sketched in Figure 6, with $K (d)$ in blue.

A two-dimensional graph displays two linear functions. The x-axis is labeled 'd' and ranges from 0 to 160, with major tick marks every 10 units. The y-axis is labeled with a dollar sign and ranges from 0 to 120, with major tick marks every 10 units. The orange line represents the function K(d) = 0.59d + 20. The blue line represents the function M(d) = 0.63d + 16. The two lines intersect at the point (100, 80), which is explicitly marked on the graph. — Figure 6

To find the intersection, we set the equations equal and solve:

\begin{array}{l} K (d) = M (d) \\ 0.59 d + 20 = 0.63 d + 16 \\ 4 = 0.04 d \\ 100 = d \\ d = 100 \end{array}

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that $K (d)$ is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is $d > 100.$

We can use the same problem strategies that we would use for any type of function.
When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See Example 1.
Draw a diagram, where appropriate. See Example 2 and Example 3.
Check for reasonableness of the answer.
Linear models may be built by identifying or calculating the slope and using the y-intercept.
The x-intercept may be found by setting $y = 0,$ which is setting the expression $m x + b$ equal to 0.
The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See Example 4.
A graph of the system may be used to identify the points where one line falls below (or above) the other line.

Verbal

Exercise 1

Explain how to find the input variable in a word problem that uses a linear function.

Solution

Determine the independent variable. This is the variable upon which the output depends.

Exercise 2

Explain how to find the output variable in a word problem that uses a linear function.

Exercise 3

Explain how to interpret the initial value in a word problem that uses a linear function.

Solution

To determine the initial value, find the output when the input is equal to zero.

Exercise 4

Explain how to determine the slope in a word problem that uses a linear function.

Algebraic

Exercise 5

Find the area of a parallelogram bounded by the y-axis, the line $x = 3,$ the line $f (x) = 1 + 2 x,$ and the line parallel to $f (x)$ passing through $(2, 7) .$

Solution

6 square units

Exercise 6

Find the area of a triangle bounded by the x-axis, the line $f (x) = 12 - \frac{1}{3} x,$ and the line perpendicular to $f (x)$ that passes through the origin.

Exercise 7

Find the area of a triangle bounded by the y-axis, the line $f (x) = 9 - \frac{6}{7} x,$ and the line perpendicular to $f (x)$ that passes through the origin.

Solution

20.012 square units

Exercise 8

Find the area of a parallelogram bounded by the x-axis, the line $g (x) = 2,$ the line $f (x) = 3 x,$ and the line parallel to $f (x)$ passing through $(6, 1) .$

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped to 4,700. Assume this trend continues.

Exercise 9

Predict the population in 2016.

Solution

2,300

Exercise 10

Identify the year in which the population will reach 0.

For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.

Exercise 11

Predict the population in 2016.

Solution

64,170

Exercise 12

Identify the year in which the population will reach 75,000.

For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.

Exercise 13

Find the linear function that models the town’s population $P$ as a function of the year, $t,$ where $t$ is the number of years since the model began.

Solution

$P (t) = 75, 000 + 2,500 t$

Exercise 14

Find a reasonable domain and range for the function $P .$

Exercise 15

If the function $P$ is graphed, find and interpret the x- and y-intercepts.

Solution

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

Exercise 16

If the function $P$ is graphed, find and interpret the slope of the function.

Exercise 17

When will the output reached 100,000?

Solution

Ten years after the model began.

Exercise 18

What is the output in the year 12 years from the onset of the model?

For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.

Exercise 19

Find the linear function that models the baby’s weight $W$ as a function of the age of the baby, in months, $t .$

Solution

$W (t) = 0. 5 t + 7.5$

Exercise 20

Find a reasonable domain and range for the function $W$ .

Exercise 21

If the function $W$ is graphed, find and interpret the x- and y-intercepts.

Solution

$(- 15, 0)$ : The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. $(0, 7.5)$ : The baby weighed 7.5 pounds at birth.

Exercise 22

If the function W is graphed, find and interpret the slope of the function.

Exercise 23

When did the baby weigh 10.4 pounds?

Solution

At age 5.8 months.

Exercise 24

What is the output when the input is 6.2? Interpret your answer.

For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were afflicted.

Exercise 25

Find the linear function that models the number of people inflicted with the common cold $C$ as a function of the year, $t .$

Solution

$C (t) = 12, 025 - 205 t$

Exercise 26

Find a reasonable domain and range for the function $C .$

Exercise 27

If the function $C$ is graphed, find and interpret the x- and y-intercepts.

Solution

$(58.7, 0)$ : In roughly 59 years, the number of people inflicted with the common cold would be 0. $(0,12,025)$ : Initially there were 12,025 people afflicted by the common cold.

Exercise 28

If the function $C$ is graphed, find and interpret the slope of the function.

Exercise 29

When will the number of people afflicted with the common cold reach 0?

Solution

2064

Exercise 30

In what year will the number of people afflicted with the common cold be 9,700?

Graphical

For the following exercises, use the graph in Figure 7, which shows the profit, $y,$ in thousands of dollars, of a company in a given year, $t,$ where $t$ represents the number of years since 1980.

Graph of a line from (15, 150) to (25, 130). — Figure 7

Exercise 31

Find the linear function $y,$ where $y$ depends on $t,$ the number of years since 1980.

Solution

$y = - 2 t +180$

Exercise 32

Find and interpret the y-intercept.

Exercise 33

Find and interpret the x-intercept.

Solution

In 2070, the company’s profit will be zero.

Exercise 34

Find and interpret the slope.

For the following exercises, use the graph in Figure 8, which shows the profit, $y,$ in thousands of dollars, of a company in a given year, $t,$ where $t$ represents the number of years since 1980.

Graph of a line from (15, 150) to (25, 450). — Figure 8

Exercise 35

Find the linear function $y,$ where $y$ depends on $t,$ the number of years since 1980.

Solution

$y = 30 t - 300$

Exercise 36

Find and interpret the y-intercept.

Exercise 37

Find and interpret the x-intercept.

Solution

(10, 0) In 1990, the profit earned zero profit.

Exercise 38

Find and interpret the slope.

Numeric

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 2. Assume that the house values are changing linearly.

Table 2 ..
Year	Mississippi	Hawaii
1950	$25,200	$74,400
2000	$71,400	$272,700

Exercise 39

In which state have home values increased at a higher rate?

Solution

Hawaii

Exercise 40

If these trends were to continue, what would be the median home value in Mississippi in 2010?

Exercise 41

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.)

Solution

During the year 1933

For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 3. Assume that the house values are changing linearly.

Table 3 ..
Year	Indiana	Alabama
1950	$37,700	$27,100
2000	$94,300	$85,100

Exercise 42

In which state have home values increased at a higher rate?

Exercise 43

If these trends were to continue, what would be the median home value in Indiana in 2010?

Solution

$105,620

Exercise 44

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 45

In 2004, a school population was 1,001. By 2008 the population had grown to 1,697. Assume the population is changing linearly.

ⓐ How much did the population grow between the year 2004 and 2008?
ⓑ How long did it take the population to grow from 1,001 students to 1,697 students?
ⓓ What was the population in the year 2000?
ⓔ Find an equation for the population, $P,$ of the school t years after 2000.
ⓕ Using your equation, predict the population of the school in 2011.

Solution

ⓐ 696 people
ⓑ 4 years
ⓓ 305 people
ⓔ $P (t) = 305 + 174 t$
ⓕ 2,219 people

Exercise 46

In 2003, a town’s population was 1,431. By 2007 the population had grown to 2,134. Assume the population is changing linearly.

ⓐ How much did the population grow between the year 2003 and 2007?
ⓑ How long did it take the population to grow from 1,431 people to 2,134 people?
ⓓ What was the population in the year 2000?
ⓔ Find an equation for the population, $P$ of the town $t$ years after 2000.
ⓕ Using your equation, predict the population of the town in 2014.

Exercise 47

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 for voice and video calling. 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.

ⓐ Find a linear equation for the monthly cost of the cell plan as a function of x, the number of monthly minutes used.
ⓑ Interpret the slope and y-intercept of the equation.

Solution

ⓐ $C (x) = 0.15 x + 10$
ⓑ The flat monthly fee is $10 and there is an additional $0.15 fee for each additional minute used
ⓒ $113.05

Exercise 48

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.

ⓐ Find a linear equation for the monthly cost of the data plan as a function of $x$ , the number of MB used.
ⓑ Interpret the slope and y-intercept of the equation.

Exercise 49

In 1990, 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.

ⓐ Find a formula for the moose population, P since 1991.
ⓑ What does your model predict the moose population to be in 2003?

Solution

ⓐ $P (t) = 190 t + 4,360$
ⓑ 6,640 moose

Exercise 50

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 2003.

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

Exercise 51

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.

ⓐ Give a linear equation for the remaining federal helium reserves, $R,$ in terms of $t,$ the number of years since 2010.
ⓑ In 2015, what will the helium reserves be?

Solution

ⓐ $R (t) = 16 - 2.1 t$
ⓑ 5.5 billion cubic feet

Exercise 52

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:

ⓐ Give a linear equation for the remaining oil reserves, $R,$ in terms of $t,$ the number of years since now.
ⓑ Seven years from now, what will the oil reserves be?

Exercise 53

You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

Solution

More than 133 minutes

Exercise 54

You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable?

Exercise 55

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?

Solution

More than $42,857.14 worth of jewelry

Exercise 56

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

Option A: Base salary of $14,000 a year with a commission of 10% of your sales
Option B: Base salary of $19,000 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?

Exercise 57

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

Option A: Base salary of $20,000 a year with a commission of 12% of your sales
Option B: Base salary of $26,000 a year with a commission of 3% of your sales

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

Solution

$66,666.67

Exercise 58

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?