2.3: Models and Applications
- Simplify Rates and Unit Rates
- Solve Linear Direct Variation Applications
- Set up a linear equation to solve a real-world application.
- Use a formula to solve a real-world application.
Rates
We now introduce the concept of rate , a special type of ratio.
A rate is a quotient of two measurements with different units.
The physical interpretation of a rate in terms of its units is an important skill.
An automobile travels 224 miles on 12 gallons of gasoline. Express the ratio distance traveled to gas consumption as a fraction reduced to lowest terms. Write a short sentence explaining the physical significance of your solution. Include units in your description.
Solution
Place miles traveled over gallons of gasoline consumed and reduce.
\[ \begin{aligned} \frac{224 \text{ mi}}{12 \text{ gal}} = \frac{56 \cdot \textcolor{red}{4} \text{ mi}}{3 \cdot \textcolor{red}{4} \text{ gal}} ~ & \textcolor{red}{ \text{ Factor.}} \\ = \frac{56 \cdot \cancel{ \textcolor{red}{4}} \text{ mi}}{3 \cdot \cancel{ \textcolor{red}{4}} \text{ gal}} ~ & \textcolor{red}{ \text{ Cancel common factor.}} \\ = \frac{56 \text{ mi}}{3 \text{ gal}} \end{aligned}\nonumber \]
Thus, the rate is 56 miles to 3 gallons of gasoline. In plain-speak, this means that the automobile travels 56 miles on 3 gallons of gasoline.
Unit Rates
When making comparisons, it is helpful to have a rate in a form where the denominator is 1. Such rates are given a special name.
A unit rate is a rate whose denominator is 1.
Herman drives 120 miles in 4 hours. Find his average rate of speed.
Solution
Place the distance traveled over the time it takes to drive that distance.
\[ \begin{aligned} \frac{120 \text{ miles}}{4 \text{ hours}} = \frac{30 \text{ miles}}{1 \text{ hour}} ~ & \textcolor{red}{ \text{ Divide: } 120/4 = 30.} \\ = 30 \text{ miles/hour} \end{aligned}\nonumber \]
Hence, Herman’s average rate of speed is 30 miles per hour.
Aditya works 8.5 hours and receives $95 in pay. What is his hourly pay rate?
Solution
Let’s place money earned over hours worked to get the following rate:
\[ \frac{95 \text{ dollars}}{8.5 \text{ hours}}\nonumber \]
We will get a much better idea of Aditya’s salary rate if we express the rate with a denominator of 1. To do so, divide. Push the decimal in the divisor to the far right, then move the decimal an equal number of places in the dividend. As we are dealing with dollars and cents, we will round our answer to the nearest hundredth.
Because the test digit is greater than or equal to 5, we add 1 to the rounding digit and truncate; i.e., 95/8.5 ≈ 11.18. Hence,
\[ \begin{aligned} \frac{95 \text{ dollars}}{8.5 \text{ hours}} = \frac{11.18 \text{ dollars}}{1 \text{ hour}} ~ & \textcolor{red}{ \text{ Divide: } 95/8.5 \approx 11.18.} \\ = 11.18 \text{ dollars/hour.} \end{aligned}\nonumber \]
That is, his pay rate is 11.18 dollars per hour.
One automobile travels 422 miles on 15 gallons of gasoline. A second automobile travels 354 miles on 13 gallons of gasoline. Which automobile gets the better gas mileage?
Solution
Decimal division (rounded to the nearest tenth) reveals the better gas mileage.
|
In the case of the first automobile, we get the following rate: \[ \frac{422 \text{ mi}}{15 \text{gal}}\nonumber \] Divide.
To the nearest tenth, 28.1. |
In the case of the second autombile, we get the following rate: \[ \frac{354 \text{ mi}}{13 \text{ gal}}\nonumber \] Divide.
To the nearest tenth, 27.2. |
In the case of the first automobile, the mileage rate is 28.1 mi/1 gal, which can be read “28.1 miles per gallon.” In the case of the second automobile, the mileage rate is 27.2 mi/1 gal, which can be read “27.2 miles per gallon.” Therefore, the first automobile gets the better gas mileage.
Solving Linear Direct Variation Applications
A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.
Nicole’s earnings can be found by multiplying her sales by her commission. The formula \(e=0.16s\) tells us her earnings, \(e\), come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 5.8.1 .
| \(s\), sales price | \(e=0.16s\) | Interpretation |
|---|---|---|
| $9,200 | \(e=0.16(9,200)=1,472\) | A sale of a $9,200 vehicle results in $1472 earnings. |
| $4,600 | \(e=0.16(4,600)=736\) | A sale of a $4,600 vehicle results in $736 earnings. |
| $18,400 | \(e=0.16(18,400)=2,944\) | A sale of a $18,400 vehicle results in $2944 earnings. |
Table 5.8.1
Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.
Figure 5.8.1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula \(y=kx\) is used for direct variation. The value \(k\) is a nonzero constant greater than zero and is called the constant of variation . In this case, \(k=0.16\) and \(n=1\). We saw functions like this one when we discussed power functions.
A General Note: DIRECT VARIATION
If \(x\) and \(y\) are related by an equation of the form
\(y=kx\)
then we say that the relationship is direct variation and \(y\) varies directly with, or is proportional to \(x\). In direct variation relationships, there is a nonzero constant ratio \(k=\dfrac{y}{x}\), where \(k\) is called the constant of variation , which help defines the relationship between the variables.
Given a description of a direct variation problem, solve for an unknown.
- Identify the units for the variable, \(x\),and the units for the variable, \(y\).
- Determine the constant of variation. You may need to divide \(y\) by the specified power of \(x\) to determine the constant of variation.
- Use the constant of variation to write an equation for the relationship.
- Substitute known values into the equation to find the unknown.
Example 5:
Solving a Direct Variation Problem
The number of kilometers \(y\) varies directly with the number of miles \(x\). If \(y=100\) km, then \(x=60\) mi, find the number of miles \(500\) that is equivalent to kilometers.
Solution
The general linear formula for direct variation is \(y=kx\). The constant can be found by dividing \(y\) by \(x\).
\(k=\dfrac{y}{x}\)
\(=\dfrac{100}{60}\)
\(=\dfrac{5}{3}\)
Now use the constant to write an equation that represents this relationship.
\(y=\dfrac{5}{3} x\)
Substitute \(y=500\) and solve for \(x\).
\(500=\dfrac{5}{3} x\)
\(1500=5x\)
\(x =\dfrac{1500}{5}\)
\(x =300\)
Answer: 500 kilometers is equivalent to 300 miles.
Q&A
Do the graphs of all linear direct variation equations pass through the origin?
YES. Linear Direct variation equations are always of the form \(y=kx\), since \(0 = k*0\) for any constant of variation \(k\), all of these graphs pass through \((0,0)\).
Example 6:
The mass of an object in kilograms varies directly with the weight in pounds. If \(y=100\) kg, then \(x=220\) lbs, find the mass in kilograms of an object that weights 55 lbs.
Solution
\( k = \frac{100}{220}\)
\( k = \frac{5}{11}\)
\( y = \frac{5}{11} x \)
since \(x = 55 \)
\( y = \frac{5}{11}(55) \)
\( y = 25 \)
Answer
An object that weights 55 lbs has a mass of 25 kilograms.
Using Rates and Direct Variation in a Formula
It takes Andrew \(30\; min\) to drive to work in the morning. He drives home using the same route, but it takes \(10\; min \) longer, and he averages \(10\; mi/h\) less than in the morning. How far does Andrew drive to work?
Solution
This is a distance problem, so we can use the formula \(d =rt\), where distance equals rate multiplied by time. Note that when rate is given in \(mi/h\), time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.
First, we identify the known and unknown quantities. Andrew’s morning drive to work takes \(30\; min\), or \(12\; h\) at rate \(r\). His drive home takes \(40\; min\), or \(23\; h\), and his speed averages \(10\; mi/h\) less than the morning drive. Both trips cover distance \(d\). A table, such as Table \(\PageIndex{2}\), is often helpful for keeping track of information in these types of problems.
| \(d\) | \(r\) | \(t\) | |
|---|---|---|---|
| To Work | \(d\) | \(r\) | \(12\) |
| To Home | \(d\) | \(r−10\) | \(23\) |
Write two equations, one for each trip.
\[d=r\left(\dfrac{1}{2}\right) \qquad \text{To work} \nonumber\]
\[d=(r-10)\left(\dfrac{2}{3}\right) \qquad \text{To home} \nonumber\]
As both equations equal the same distance, we set them equal to each other and solve for \(r\).
\[\begin{align*} r\left (\dfrac{1}{2} \right )&= (r-10)\left (\dfrac{2}{3} \right )\\ \dfrac{1}{2r}&= \dfrac{2}{3}r-\dfrac{20}{3}\\ \dfrac{1}{2}r-\dfrac{2}{3}r&= -\dfrac{20}{3}\\ -\dfrac{1}{6}r&= -\dfrac{20}{3}\\ r&= -\dfrac{20}{3}(-6)\\ r&= 40 \end{align*}\]
We have solved for the rate of speed to work, \(40\; mph\). Substituting \(40\) into the rate on the return trip yields \(30 mi/h\). Now we can answer the question. Substitute the rate back into either equation and solve for \(d\).
\[\begin{align*}d&= 40\left (\dfrac{1}{2} \right )\\ &= 20 \end{align*}\]
The distance between home and work is \(20\; mi\).
Analysis
Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for \(r\).
\[\begin{align*} r\left (\dfrac{1}{2} \right)&= (r-10)\left (\dfrac{2}{3} \right )\\ 6\times r\left (\dfrac{1}{2} \right)&= 6\times (r-10)\left (\dfrac{2}{3} \right )\\ 3r&= 4(r-10)\\ 3r&= 4r-40\\ r&= 40 \end{align*}\]
Building Linear Models
When modeling scenarios with linear equations and solving problems involving quantities with a constant rate of change , we typically follow the 5 step method.
We list the strategy here, and then use it to solve some problems.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for and choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem, then translate into an algebraic equation.
- Solve the equation using algebraic techniques.
- Check the answer in the problem and make sure it makes sense. Clearly state your result using appropriate units.
Example 8:
Emily is a college student who plans to spend a summer in Seattle. She has saved $3,600 for her trip and anticipates spending $400 each day on food, lodging and activities. How many days can Emily remain in Seattle before her money runs out?
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.
- Dependent: M, money remaining, in dollars
- Independent: t, time, in days
So, the amount of money M, remaining depends on the number of days t
We can also identify the initial value and the rate of change.
- Initial Value: She saved $3,600, so 3600 is the initial value for M.
- Rate of Change: She anticipates spending $400 each day, so –400 is the slope.
Notice that the unit of dollars per day 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 day.
Since this student is spending the same amount of money each day, the rate of change is constant ,
so we can start with the linear model \(M =mt+b\).
Then we can substitute \(m= -400\) for the slope and \(b=3600\) for the initial value.
\(M =-400t+3600\)
Solution
To find the t-intercept, we set the amount of money to zero, and solve for the time.
0=−400t+3600
400t = 3600
t=9
Answer :
The t-intercept is \((9,0)\) we say that Emily will have no money left after 9 days.
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.
Data can be used to construct functions that model real-world applications. Once an equation that fits given data is determined, we can use the equation to make certain predictions; this is called mathematical modeling 35 .
Example 9:
Suppose 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, \(y=mx+b\)
Solution
\(y = −250x+1000\)
Now we can set y equal to 0, and solve for x to find the x-intercept.
\(0=−250x+1000\)
\(250x = 1000\)
\(x=4\)
The \(x\)-intercept is the number of months it takes her to reach a balance of $0.
Answer :
The x-intercept is 4 months, so it will take Hannah four months to pay off her loan.
Using Two Given Pairs of Values 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 model. We might sometimes instead be asked to evaluate the linear model at a given value or set the equation of the linear model equal to a specified value.
Example 10:
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:
- Independent: t, years since 2004
- Dependent: P, 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 number of years that would produce a population 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 population/change in time
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.
\(m= \frac {8100 - 6200 } { 5 - 0 }\)
= 380 people per year
We already know the y-intercept of the line, so we can immediately write the equation:
P=380t+6200
To predict the population in 2013, we evaluate our equation at t=9.
P=380(9)+6,200
P=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=15000 and solve for t.
15000=380t+6200
8800 = 380t
t≈23.158
Answer : Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.
Modeling Cost Applications
Data can be used to construct functions that model real-world applications. Once an equation that fits given data is determined, we can use the equation to make certain predictions; this is called mathematical modeling 35 .
Example 11:
The cost of a daily truck rental is \($48.00\), plus an additional \($0.45\) for every mile driven. Write a function that gives the cost of the daily truck rental and use it to determine the total cost of renting the truck for a day and driving it \(60\) miles.
Solution
The total cost of the truck rental depends on the number of miles driven. If we let \(x\) represent the number of miles driven, then \(0.45x\) represents the variable cost of renting the truck. Use this and the fixed cost, \($48.00\), to write a function that models the total cost,
\(C = 0.45 x + 48\)
Use this function to calculate the cost of the rental when \(x = 60\) miles.
\(\begin{aligned} C & = 0.45 ( 60 ) + 48 \\ & = 27 + 48 \\ & = 75 \end{aligned}\)
Answer :
The total cost of renting the truck for the day and driving it \(60\) miles would be \($75\).
We can use the model \(C = 0.45x + 48\) to answer many more questions. For example, how many miles can be driven to keep the cost of the rental at most \($66\)? To answer this question, set up an inequality that expresses the cost less than or equal to \($66\).
\(\begin{aligned} C & \leq \$ 66 \\ 0.45 x + 48 & \leq 66 \end{aligned}\)
Solve for \(x\) to determine the number of miles that can be driven.
\(\begin{aligned} 0.45 x + 48 & \leq 66 \\ 0.45 x & \leq 18 \\ x & \leq 40 \end{aligned}\)
To limit the rental cost to \($66\), the truck can be driven \(40\) miles or less.
Modeling Depreciation of Value
Example 12:
A company purchased a new piece of equipment for \($12,000\). Four years later it was valued at \($9,000\) dollars. Use this data to construct a linear function that models the value of the piece of equipment over time.
Solution
The value of the item depends on the number of years after it was purchased. Therefore, the age of the piece of equipment is the independent variable. Use ordered pairs where the \(x\)-values represent the age and the \(y\)-values represent the corresponding value.
\((age, value)\)
From the problem, we can determine two ordered pairs. Purchased new \((age = 0)\), the item cost \($12,000\), and \(4\) years later the item was valued at \($9,000\). Therefore, we can write the following two \((age, value)\) ordered pairs:
\(( 0,\:12000 ) \quad \text { and } \quad ( 4,\: 9000 )\)
Use these two ordered pairs to construct a linear model. Begin by finding the slope \(m\).
\(\begin{aligned} m & = \frac { y _ { 2 } - y _ { 1 } } { x _ { 2 } - x _ { 1 } } \\ & = \frac { 9000 - 12000 } { 4 - 0 } \\ & = \frac { - 3000 } { 4 } \\ & = - 750 \end{aligned}\)
Here we have \(m = −750\). The ordered pair \((0, 12000)\) gives the \(y\)-intercept; therefore, \(b = 12000\).
\(\begin{array} { l } { y = m x + b } \\ { y = - 750 x + 12000 } \end{array}\)
Lastly, write this model as a function which gives the value of the piece of equipment over time. Choose the function name \(V\), for value, and the variable \(t\) instead of \(x\) to represent time in years.
\(V = - 750 t + 12000\)
Answer :
\(V = - 750 t + 12,000\)
The equation \(V = −750t + 12,000\) called a linear depreciation model 36 . It uses a linear equation to expresses the declining value of an item over time. Using this to determine the value of the item between the given data points is called interpolation 37 . For example, we can use the equation to determine the value of the item where \(t = 2\),
\(\begin{aligned} V & = - 750 ( 2 ) + 12,000 \\ & = 10,500 \end{aligned}\)
This shows that the item was worth \($10,500\) two years after it was purchased. Using this model to predict the value outside the given data points is called extrapolation 38 . For example, we can use the equation to determine the value of the item when \(t = 10\):
\(\begin{aligned} V & = - 750 ( 10 ) + 12,000 \\ & = - 7,500 + 12,000 \\ & = 4,500 \end{aligned}\)
The model predicts that the piece of equipment will be worth \($4,500\) ten years after it is purchased.
Business Models
In a business application, revenue results from the sale of a number of items. For example, if an item can be sold for \($150\) and we let \(n\) represent the number of units sold, then we can form the following revenue equation 39 :
\(R=150n\)
Use this function to determine the revenue generated from selling \(n=100\) units,
\(R = 150 ( 100 ) = 15,000\)
The function shows that the revenue generated from selling \(100\) items is \($15,000\). Typically, selling items does not represent the entire story. There are a number of costs associated with the generation of revenue. For example, if there is a one-time set up fee of \($5,280\) and each item cost \($62\) to produce, then we can form the following cost equation 40 :
\(C = 62 n + 5,280\)
Here \(n\) represents the number of items produced. Use this to determine the cost associated with producing \(n = 100\) units:
\(C = 62 ( 100 ) + 5,280 = 11,480\)
This shows that the cost associated with producing 100 items is $11,480. Profit is revenue less costs:
\(\begin{aligned} \color{Cerulean} { Profit } & = \color {Cerulean} { Revenue -Cost } \\ & = 15,000 - 11,480 \\ & = 3,520 \end{aligned}\)
Therefore, the profit generated by producing and selling \(100\) items is \($3,520\). In general, given a revenue equation\(R\) and a cost equation \(C\), we can form a profit equation 41 by subtracting as follows:
\(P = R - C \)
Example 13:
The cost in dollars of producing n items is given by the formula \(C = 62n + 5,280\). The revenue in dollars is given by \(R = 150n\), where \(n\) represents the number items sold. Write an equation that gives the profit generated by producing and selling n items. Use this equation to determine how many items must be produced and sold in order to earn a profit of at least \($7,000\).
Solution
Obtain the profit equation by subtracting the cost equation from the revenue equation.
\(\begin{aligned} P & = R - C \\ & = 150 n - ( 62 n + 5,280 ) \\ & = 150 n - 62 n - 5,280 \\ & = 88 n - 5,280 \end{aligned}\)
Therefore, \(P (n) = 88n + 5,280\) models the profit. To determine the number of items that must be produced and sold to profit at least \($7,000\), solve the following:
\(\begin{aligned} P & \geq 7,000 \\ 88 n - 5,280 & \geq 7,000 \\ 88 n & \geq 12,280 \\ n & \geq 139.5 \end{aligned}\)
Round up because the number of units produced and sold must be an integer. To see this, calculate the profit where \(n\) is \(139\) and \(140\) units.
\(\begin{array} { l } { P = 88 ( 139 ) - 5,280 = 6,952 } \\ { P = 88 ( 140 ) - 5,280 = 7,040 } \end{array}\)
Answer :
\(140\) or more items must be produced and sold in order to earn a profit of at least \($7,000\).
Sometimes the costs exceed the revenue, in which case, the profit will be negative. For example, use the profit equation of the previous example, \(P = 88n − 5,280\), to calculate the profit generated where \(n = 50\).
\(P = 88 ( 50 ) - 5,280 = - 880\)
This indicates that when \(50\) units are produced and sold the corresponding profit is a loss of \($880\).
It is often important to determine how many items must be produced and sold to break even. To break even means to neither have a gain nor a loss; in this case, the profit will be equal to zero. To determine the breakeven point 42 , set the profit function equal to zero and solve:
\(\begin{aligned} P & = 88 n - 5,280 \\ 0 & = 88 n - 5,280 \\ 5,280 & = 88 n \\ 60 & = n \end{aligned}\)
Therefore, \(60\) items must be produced and sold to break even.
You Try it.
Custom t-shirts can be sold for \($6.50\) each. In addition to an initial setup fee of \($120\), each t-shirt cost \($3.50\) to produce.
- Write an equation that models the revenue and an equation that models the cost.
- Determine the equation that models the profit and use it to determine the profit from producing and selling \(150\) t-shirts.
- Calculate the number of t-shirts that must be sold to break even.
- Answer
-
a. Revenue: \(R = 6.50x\); cost: \(C = 3.50x + 120\);
b. profit: \(P = 3x - 120\); profit from producing and selling 150 t-shirts: \($330\)
c. \(40\)
Key Concepts
- A rate is a quotient of two measurements with different units.
- There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the \(d = rt\) formula. See Example .A linear equation can be used to solve for an unknown in a number problem. See Example .
- Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See Example .
- Given two points we can find the equation of a line.
- To find an equation of a line, first use the given information to determine the slope. Then use the slope and a point on the line to find the equation using point-slope form.
- To construct a linear equation that models a real-world application, first identify the dependent and independent variables. Next, find two ordered pairs that describe the given situation. Use these two ordered pairs to construct a linear function by finding the slope and \(y\)-intercept.