5.7: Problem Solving Strategies: Use a Linear Model
Learning Objectives
 Read and understand given problem situations.
 Develop and apply the strategy: use a linear model.
 Plan and compare alternative approaches to solving problems.
 Solve realworld problems using selected strategies as part of a plan.
Introduction
In this chapter, we have been estimating values using straight lines. When we use linear interpolation, linear extrapolation or predicting results using a line of best fit, it is called linear modeling. In this section, we will look at a few examples where data sets occurring in realworld problems can be modeled using linear relationships. From previous sections remember our problem solving plan:.
Step 1
Understand the problem
Read the problem carefully. Once the problem is read, list all the components and data that are involved. This is where you will be assigning your variables
Step 2
Devise a plan – Translate
Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or construct a table as a start to solving your problem.
Step 3
Carry out the plan – Solve
This is where you solve the equation you came up with in Step 2.
Step 4
Look – Check and Interpret
Check to see if you used all your information and that the answer makes sense.
Example 1
Dana heard something very interesting at school. Her teacher told her that if you divide the circumference of a circle by its diameter you always get the same number. She tested this statement by measuring the circumference and diameter of several circular objects. The following table shows her results.
From this data, estimate the circumference of a circle whose diameter is 12 inches. What about 25 inches? 60 inches?
Object  Diameter (inches)  Circumference (inches) 

Table  53  170 
Soda can  2.25  7.1 
Cocoa tin  4.2  12.6 
Plate  8  25.5 
Straw  .25  1.2 
Propane tank  13.3  39.6 
Hula Hoop  34.25  115 
Solution
Let’s use the problem solving plan.
Step 1
We define our variables.
\begin{align*}x =\end{align*}
\begin{align*}y =\end{align*}
We want to know the circumference when the diameter is 12, 25 or 60 inches.
Step 2 We can find the answers either by using the line of best fit or by using linear interpolation or extrapolation. We start by drawing the scatter plot of the data.
Step 3 Line of best fit
Estimate a line of best fit on the scatter plot.
Find the equation using points (.25, 1.2) and (8, 25.5).
\begin{align*}\text{Slope} \ m & = \frac{25.5  1.2} {8  .25} = \frac{24.3} {7.75} = 3.14\\
y & = 3.14x + b\\
1.2 & = 3.14 (.25) + b \Rightarrow b = 0.42\\
\text{Equation} \ y & = 3.14x + 0.42\end{align*}
\begin{align*}\text{Diameter} & = 12 \ inches \Rightarrow y = 3.14 (12) + 0.42 = \underline{38.1\ inches}\\
\text{Diameter} & = 25 \ inches \Rightarrow y = 3.14 (25) + 0.42 = \underline{78.92\ inches}\\
\text{Diameter} & = 60 \ inches \Rightarrow y = 3.14 (60) + 0.42 = \underline{188.82\ inches}\end{align*}
In this problem the \begin{align*}\text{slope} = 3.14\end{align*}
You are probably more familiar with the formula \begin{align*}C = \pi \cdot d\end{align*}
Note: The calculator gives the line of best fit as \begin{align*}y = 3.25x 0.57\end{align*}
Step 4 Check and Interpret
The circumference of a circle is \begin{align*}\pi d\end{align*}
The reason the line of best fit method works the best is that the data is very linear. All the points are close to the straight line but there is some slight measurement error. The line of best fit averages the error and gives a good estimate of the general trend.
Note: The linear interpolation and extrapolation methods give estimates that aren't as accurate because they use only two points in the data set. If there are measurement errors in the points that are being used, then the estimates will lose accuracy. Normally, it is better to compute the line of best fit with a calculator or computer.
Example 2
A cylinder is filled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of the cylinder and measurements are taken at two second intervals. The table below shows the height of the water level in the cylinder at different times.
a) Find the water level at 15 seconds.
b) Find the water level at 27 seconds.
Water Level in Cylinder at Various Times
Time (seconds)  Water level (cm) 

0.0  73 
2.0  63.9 
4.0  55.5 
6.0  47.2 
8.0  40.0 
10.0  33.4 
12.0  27.4 
14.0  21.9 
16.0  17.1 
18.0  12.9 
20.0  9.4 
22.0  6.3 
24.0  3.9 
26.0  2.0 
28.0  0.7 
30.0  0.1 
Solution
Let’s use the problem solving plan.
Step 1
Define our variables
\begin{align*}x=\end{align*}
\begin{align*}y =\end{align*}
We want to know the water level at time 15, 27 and 5 seconds.
Step 2 We can find the answers either by using the line of best fit or by using linear interpolation or extrapolation. We start by drawing the scatter plot of the data.
Step 3 Method 1 Line of best fit
Draw an estimate of the line of best fit on the scatter plot. Find the equation using points (6, 47.2) and (24, 3.9).
\begin{align*}\text{Slope} & & m & = \frac{3.9  47.2} {24  6} = \frac{43.3} {18} =  2.4\\
& & y & = 2.4x + b\\
& & 47.2 & = 2.4(6) + b \Rightarrow b = 61.6\\
\text{Equation} & & y & = 2.4 x + 61.6\end{align*}
\begin{align*}\text{Time} & = 15 \ seconds \Rightarrow y = 2.4 (15) + 61.6 = \underline{25.6 \ cm}\\
\text{Time} & = 27 \ seconds \Rightarrow y = 2.4 (27) + 61.6 = \underline{3.2 \ cm}\end{align*}
The line of best fit does not show us accurate estimates for the height. The data points do not appear to fit a linear trend so the line of best fit is close to very few data points.
Method 2: Linear interpolation or linear extrapolation.
We use linear interpolation to find the water level for the times 15 and 27 seconds, because these points are between the points we know.
\begin{align*}\text{Time} = 15 \ seconds \end{align*}
Connect points (14, 21.9) and (16, 17.1) and find the equation of the straight line.
\begin{align*} m = \frac{17.1  21.9} {16  14} = \frac{4.8} {2} = 2.4 \ y = 2.4x + b \Rightarrow 21.9 = 2.4 (14) + b \Rightarrow b = 55.5\end{align*}
Equation \begin{align*}y = 2.4x + 55.5\end{align*}
Plug in \begin{align*}x=15\end{align*}
\begin{align*}\text{Time} = 27 \ seconds\end{align*}
Connect points (26, 2) and (28, 0.7) and find the equation of the straight line.
\begin{align*} m & = \frac{0.7  2} {28  26} = \frac{1.3} {2} = .65\\
y & = .65x + b \Rightarrow 2 =  .65 (26) + b \Rightarrow b = 18.9\end{align*}
Equation \begin{align*}y= .65x = 18.9\end{align*}
Plug in \begin{align*}x = 27\end{align*}
We use linear extrapolation to find the water level for time 5 seconds because this point is smaller than the points in our data set.
Step 4 Check and Interpret
In this example, the linear interpolation and extrapolation method gives better estimates of the values that we need to solve the problem. Since the data is not linear, the line of best fit is not close to many of the points in our data set. The linear interpolation and extrapolation methods give better estimates because we do not expect the data to change greatly between the points that are known.
Lesson Summary
 Using linear interpolation, linear extrapolation or prediction using a line of best fit is called linear modeling.
 The four steps of the problem solving plan are:
 Understand the problem
 Devise a plan – Translate
 Carry out the plan – Solve
 Look – Check and Interpret
Review Questions
The table below lists the predicted life expectancy based on year of birth (US Census Bureau).
Use this table to answer the following questions.
Birth Year  Life expectancy in years 

1930  59.7 
1940  62.9 
1950  68.2 
1960  69.7 
1970  70.8 
1980  73.7 
1990  75.4 
2000  77 
 Make a scatter plot of the data
 Use a line of best fit to estimate the life expectancy of a person born in 1955.
 Use linear interpolation to estimate the life expectancy of a person born in 1955.
 Use a line of best fit to estimate the life expectancy of a person born in 1976.
 Use linear interpolation to estimate the life expectancy of a person born in 1976.
 Use a line of best fit to estimate the life expectancy of a person born in 2012.
 Use linear extrapolation to estimate the life expectancy of a person born in 2012.
 Which method gives better estimates for this data set? Why?
The table below lists the high temperature for the first day of the month for year 2006 in San Diego, California (Weather Underground).
Use this table to answer the following questions.
Month number  Temperature (F) 

1  63 
2  66 
3  61 
4  64 
5  71 
6  78 
7  88 
8  78 
9  81 
10  75 
11  68 
12  69 
 Draw a scatter plot of the data
 Use a line of best fit to estimate the temperature in the middle of the \begin{align*}4^{th}\end{align*}
4th month (month 4.5).  Use linear interpolation to estimate the temperature in the middle of the \begin{align*}4^{th}\end{align*}
4th month (month 4.5).  Use a line of best fit to estimate the temperature for month 13 (January 2007).
 Use linear extrapolation to estimate the temperature for month 13 (January 2007).
 Which method gives better estimates for this data set? Why?
Review Answers
 Equation of line of best fit using points (1940, 62.9) and (1990, 75.4) \begin{align*}y = .25x 422.1\end{align*}
y=.25x−422.1 .  66.7 years

\begin{align*}y = .15x 224.3\end{align*}
y=.15x−224.3 , 69.0 years  71.9 years

\begin{align*}y = .29x 500.5\end{align*}
y=.29x−500.5 , 72.5 years  80.9 years

\begin{align*}y= .16x 243\end{align*}
y=.16x−243 , 78.9 years  A line of best fit gives better estimates because data is linear.
 Equation of line of best fit using points (2, 66) and (10, 75) \begin{align*}y = 1.125x + 63.75\end{align*}
y=1.125x+63.75  68.8 F

\begin{align*}y = 7x + 36\end{align*}
y=7x+36 , 67.5 F  78.4 F

\begin{align*}y = x + 57\end{align*}
y=x+57 , 70 F  Linear interpolation and extrapolation give better estimates because data is not linear.
Texas Instruments Resources
In the CK12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9615.
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