7.2: Modeling with Linear Equations
In this section, you will learn to use linear functions to model real-world applications
Read the problem carefully. Highlight important information. Identify what each variable represents in the context of the problem.
It costs $750 to manufacture 25 items, and $1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this equation to predict the cost of 100 items.
Solution
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 \(y = 10x + 500\). This equation is the linear model for this problem.
Now use the linear model 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. Write a conversion equation from Celsius to Fahrenheit and use this equation to convert 30 degrees Celsius into Fahrenheit.
Solution
Let us look at what is given.
| Celsius | Fahrenheit |
| 0 | 32 |
| 100 | 212 |
We let \(C\) = the degrees in Celsius, and let \(F\) = the degrees in Fahrenheit.
Again, 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 in this case \(F = mC + b\), where \(x\) or \(C\) represent the temperature in Celsius, and y or F the temperature in Fahrenheit.
\[ \text{slope m } = \frac{312-32}{100-0} = \frac{9}{5} \nonumber \]
The equation is \(F = \frac{9}{5}C + b\)
Substituting the point (0, 32), we get b = 32 and the conversion equation is
\[F = \frac{9}{5}C + 32 \nonumber.\]
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}
Thus, 30 degrees Celsius is equal to 86 degrees Fahrenheit.
The variable cost to manufacture a product is $10 per item and the fixed cost $2500. If \(x\) represents the number of items manufactured and \(y\) represents the total cost, write the cost function.
Solution
- 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 function is \(y = 10x + 2500\).
Assume a car depreciates by the same amount each year. Joe purchased a car in 2010 for $16,800. In 2014 it is worth $12,000. Find the linear model. Use this model to predict how much the car will be worth in 2020.
Solution:
We let \(x\) = the number of years after 2010, and let \(y\) = the cost.
Solving this problem is equivalent to finding an equation of a line that passes through the points (0, 16800) and (4, 12000).
To find the linear model for this problem, we need to find the slope.
\[ m = \frac{12000-16800}{4-0} = -1200 \nonumber \]
The slope indicates that the rate of depreciation each year is $-1200. Thus, the linear model for this problem is: \(y = -1200x + 16,800\)
Now, to find out how much the car will be worth in 2020, we need to know how many years that is from the purchase year. Since it is ten years later, \(x=10\).
\[y=-1200(10)+16,800=-12,000+16,800=4,800\]
The car will be worth $4800 in 2020.
Note: The value of the car over time follows a decreasing straight line.
The cost \(y\), in dollars, of a gym membership for \(n\) months can be described by the linear model \(y=30n+70\). What does this model tell us?
Solution
The value for \(y\) when \(n=0\) in this equation is 70, so the initial starting cost is $70. This tells us that there must be an initiation or start-up fee of $70 to join the gym.
The value for the slope, \(m\) in the equation is 30, so the cost increases by $30 each month. This tells us that the monthly membership fee for the gym is $30 a month.
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 \(b = 24.5\). We write the linear model
\[ y =0.32x + 24.5 \nonumber \]
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.
A quantity grows linearly if it grows by a constant amount for each unit of time.
Suppose in Flagstaff Arizona, the number of residents increased by 1000 people per year. If the initial population was 46,080 in 1990, can you predict the population in 2013? This is an example of linear growth because the population grows by a constant amount. We list the population in future years below by adding 1000 people for each passing year.
| 1990 | 1991 | 1992 | 1993 | 1994 | 1995 | 1996 | |
|---|---|---|---|---|---|---|---|
|
Year |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
|
Population |
46,080 |
47,080 |
48,080 |
49,080 |
50,080 |
51,080 |
52,080 |
Solution:
The population growth, y , can be modeled with a linear equation. The initial population is 46,080. The future population depends on the number of years, t , after the initial year. The model is \(y = 1000t + 46,080\). Note, we chose to use the variable t as a simple reminder that t represents time . We could continue to use the variable x , or any other letter for that matter, but t for time makes sense.
To predict the population in 2013, we identify how many years it has been from 1990 (which is year zero). So t = 23 for the year 2013.
\[y=1000(23)+46,080=69,080\]
The population of Flagstaff in 2013 will be 69,080 people.
Dora has inherited a collection of 30 antique frogs. Each year she vows to buy two frogs a month to grow the collection. This is an additional 24 frogs per year. How many frogs will she have in six years? How long will it take her to reach 510 frogs?
Solution
The initial population is 30 frogs, so b=30. The rate of change is 24 frogs per year, so m=24. The linear growth model for this problem is:
\[y = 24t + 30\] where t = time in years and y = the number of frogs
The first question asks how many frogs will Dora have in six years so, t = 6.
\[y = 24(6)+30 = 144+30 = 174\] frogs.
The second question asks for the time it will take for Dora to collect 510 frogs. So, \(y = 510\)and we will solve for t.
\[\begin{align*}510 &= 24t+30 \\ 480 &= 24t \\ 20 &= t \end{align*}\]
It will take 20 years to collect 510 antique frogs.
Note: The graph of the number of antique frogs Dora accumulates over time follows a straight line.
The number of stay-at-home fathers in Canada has been growing steadily[1]. While the trend is not perfectly linear, it is fairly linear. Use the data from 1976 and 2010 to find an explicit formula for the number of stay-at-home fathers, then use it to predict the number of stay-at-home fathers in 2020.
\(\begin{array}{|l|l|l|l|l|l|}
\hline \text { Year } & 1976 & 1984 & 1991 & 2000 & 2010 \\
\hline \text { Number of stay-at-home fathers } & 20,610 & 28,725 & 43,530 & 47,665 & 53,555 \\
\hline
\end{array}\)
- Answer
-
We let \(t\) = the number of years after 1976, and let \(y\) = the number of stay-at-home fathers.
From the table we know that 1976 corresponds to \(t=0\) and the number of stay-at-home fathers is \(y=20,610\).
From 1976 to 2010 the number of stay-at-home fathers increased by
\(53,555-20,610=32,945\)
This happened over 34 years, so the rate of change (slope) is \(32,945 / 34=969\).
\(y=969t+20,610\)
Predicting for 2020, we use \(t=44\)
\(y=969(44)+20,610=63,246\)
There will be 63,246 stay-at-home fathers in 2020.
[1] www.fira.ca/article.php?id=140