3.1: Linear Applications
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, you will learn to use linear equations to model real-world applications.
Before you get started, take this prerequisite quiz.
1. Find the slope of the line containing the points (3, 10) and (5, 40).
- Click here to check your answer
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\(15\)
If you missed this problem, review Section 1.2. (Note that this will open in a new window.)
2. Find the equation of the line containing the points (3, 10) and (5, 40). Write the equation in slope-intercept form.
- Click here to check your answer
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\(y=15x-35\)
If you missed this problem, review Section 1.3. (Note that this will open in a new window.)
Now that we have learned to solve equations and inequalities, we get to apply these ideas in a variety of real-life situations. Many scenarios follow linear trends. In this section, we'll work on creating a linear equations.
Creating Linear Equations
While each application is different, the following steps represent a general pattern.
- Read the problem carefully. Highlight important information.
- Identify the two quantities that change and define the unknowns.
- One quantity will be independent, meaning it is going to change on its own. In many applications, this quantity is time. Define this quantity as \(x\).
- Another quantity will be dependent, meaning that its value will depend on how the other quantity changes. In many applications, this is money. Define this quantity as \(y\).
- Create an equation to represent the relationship between the two variables. When applications have a starting value and a constant rate of change, these will most commonly be written in slope-intercept form, \(y=mx+b\). The y-intercept will represent the starting value and the slope will represent the rate of change.
Jadon wants to purchase a new car for $12,500. They also need to budget for expenses of gas, insurance, registration, and repairs, which they estimate to be $450 per month.
a. What will be the total cost of the car if they own it for 10 months?
b. What will be the total cost of the car if they own it for 3 years?
c. What will be the total cost of the car if they don't know how long they will own it?
Solution
a. For only 10 months, they would need to pay the purchase price once and the monthly cost 10 times. The cost would be \($12,500 + 10($450)\), which is \($17,000\).
b. For 3 years, they would need to pay the purchase price once and the monthly cost 36 (12*3) times. The cost would be \($12,500 + 36($450)\), which is \($28,700\).
c. When we don't know the time they'll own the car, we need to use variables to represent the changing quantities.
- The independent variable will be the number of months Jadon owns the car. We'll let \(x\) = the number of months.
- The cost of the car depends on the number of months they own it, so we'll let \(y\) = the cost.
Since the cost begins at $12,500 and increases $350 for every month, the equation can be written as \(y = 12500 + 450x\) or \(y = 450x + 12500\).
Note that it is also acceptable to choose other variables to represent the quantities. This same equation might be written as \(C = 12500 + 450t\), where \(t\) represents the number of months and \(C\) represents the cost.
In the problem above, $450 per month is referred to as the variable cost, which usually represents the slope in the equation. The flat charge $12,500 is referred to as the fixed cost, and usually represents the y-intercept in the equation.
Since the equation gives the total cost as the output of the equation, this is referred to as the cost function.
A taxi service charges $0.50 per mile plus a $5 flat fee.
a. What will be the cost of traveling 20 miles?
b. What is the cost function?
Solution
a. The cost of traveling 20 miles is \(5+ (0.50)(20)\) which is \($15\).
b. To find the cost function of traveling an unknown number of miles,
- Let \(x\) = distance traveled in miles
- Let \(y\) = cost in dollars
The total cost is $5 plus $0.50 for every mile traveled, so the cost function is \(y = 5 + 0.50x\).
The variable cost to manufacture a product is $10 per item and the fixed cost $2500. Define the unknowns and write the cost function.
Solution
Let \(x\) represent the number of items manufactured and let \(y\) represents the total cost
The variable cost of $10 per item tells us that \(m = 10\).
The fixed cost represents the \(y\)-intercept. So \(b = 2500\).
Therefore, the cost equation is \(y = 10x + 2500\).
It costs $750 to manufacture 25 items, and $1000 to manufacture 50 items.
a. Assuming a linear relationship holds, find the cost equation.
b. Use this function to predict the cost of 100 items.
Solution
a. We let \(x\) = the number of items manufactured, and let \(y\) = the cost.
Solving this problem is equivalent to finding an equation of a line that passes through the points (25, 750) and (50, 1000).
\[ m = \frac{1000-750}{50-25} = 10 \nonumber \]
Therefore, the partial equation is \(y = 10x + b\)
By substituting one of the points in the equation, we get \(b = 500\)
Therefore, the cost equation is \(y = 10x + 500\)
b. To find the cost of 100 items, substitute \(x = 100\) in the equation \(y = 10x + 500\)
So the cost is
\[y = 10(100) + 500 = 1500 \nonumber\]
It costs $1500 to manufacture 100 items.
The freezing temperature of water in Celsius is 0 degrees and in Fahrenheit 32 degrees. And the boiling temperatures of water in Celsius, and Fahrenheit are 100 degrees, and 212 degrees, respectively.
a. Write a conversion equation from Celsius to Fahrenheit.
b. Use this equation to convert 30 degrees Celsius into Fahrenheit.
Solution
a. Let us look at what is given.
| Celsius | Fahrenheit |
| 0 | 32 |
| 100 | 212 |
Because the directions ask us to convert from Celsius to Fahrenheit, we want Fahrenheit to be the output of our equation. Therefore, we'll let \(x\) represent Celsius and let \(y\) represent Fahrenheit.
- Note that if the directions had been written the other way (from Fahrenheit to Celsius), we would have wanted the variables to be switched.
We can now write the information given as two points of data. Solving this problem is equivalent to finding an equation of a line that passes through the points (0, 32) and (100, 212).
Since we are finding a linear relationship, we are looking for an equation \(y = mx + b\), or \(F = mC + b\) depending on what variables are defined.
\[ \text{slope m } = \frac{312-32}{100-0} = \frac{9}{5} \nonumber \]
The equation is \(F = \frac{9}{5}C + b\)
b. Substituting the point (0, 32), we get
\[F = \frac{9}{5}C + 32 \nonumber.\]
To convert 30 degrees Celsius into Fahrenheit, substitute \(C = 30\) in the equation
\begin{aligned}
&\mathrm{F}=\frac{9}{5} \mathrm{C}+32\\
&\mathrm{F}=\frac{9}{5}(30)+32=86
\end{aligned}
The population of Canada in the year 1980 was 24.5 million, and in the year 2010 it was 34 million. The population of Canada over that time period can be approximately modeled by a linear function. Let x represent time as the number of years after 1980 and let y represent the size of the population.
- Write the linear function that gives a relationship between the time and the population.
- Assuming the population continues to grow linearly in the future, use this equation to predict the population of Canada in the year 2025.
Solution
The problem can be made easier by using 1980 as the base year, that is, we choose the year 1980 as the year zero. This will mean that the year 2010 will correspond to year 30. Now we look at the information we have:
| Year | Population |
| 0 (1980) | 24.5 million |
| 30 (2010) | 34 million |
a. Solving this problem is equivalent to finding an equation of a line that passes through the points (0, 24.5) and (30, 34). We use these two points to find the slope:
\[ m = \frac{34-24.5}{30-0}=\frac{9.5}{30} = 0.32 \nonumber\]
The \(y\)-intercept occurs when \(x = 0\), so \(b = 24.5\)
\[ y =0.32x + 24.5 \nonumber \]
b. Now to predict the population in the year 2025, we let \(x=2025-1980=45\)
\begin{aligned}
&y=0.32 x+24.5\\
&y=0.32(45)+24.5=38.9
\end{aligned}
In the year 2025, we predict that the population of Canada will be 38.9 million people.
Note that we assumed the population trend will continue to be linear. Therefore if population trends change and this assumption does not continue to be true in the future, this prediction may not be accurate.