# Exercises: Linear Functions and Applications of Linear Functions (3.12)

**Exercises: Verbal Problems**

**1. Terry is skiing down a steep hill. Terry’s elevation, [latex]E(t)[/latex], in feet after [latex]t[/latex] seconds is given by [latex]E(t)=3000-70t[/latex]. Write a complete sentence describing Terry’s starting elevation and how it is changing over time.**

**Solution**

Terry starts at an elevation of [latex]3000[/latex] feet and descends [latex]70[/latex] feet per second.

**2. Jessica is walking home from a friend’s house. After [latex]2[/latex] minutes she is [latex]1.4[/latex] miles from home. Twelve minutes after leaving, she is [latex]0.9[/latex] miles from home. What is her rate in miles per hour?**

**3. A boat is [latex]100[/latex] miles away from the marina, sailing directly toward it at [latex]10[/latex] miles per hour. Write an equation for the distance of the boat from the marina after [latex]t[/latex] hours.**

**Solution**

[latex]d(t)=100-10t[/latex]

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

**5. 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.

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

**7. 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.

**Exercises: Algebraic Problems**

Instructions: For questions 8-9, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was [latex]5\text{,}900[/latex]. By 2012 the population had dropped [latex]4\text{,}700[/latex]. Assume this trend continues.

**8. Predict the population in 2016.**

**Solution**

[latex]2\text{,}300[/latex]

**9. Identify the year in which the population will reach [latex]0[/latex].**

**Exercises: Algebraic Problems**

Instructions: For questions 10-11, consider this scenario: A town’s population has been increasing at a constant rate. In 2010 the population was [latex]46\text{,}020[/latex]. By 2012 the population had increased to [latex]52\text{,}070[/latex]. Assume this trend continues.

**10.**

**Predict the population in 2016.**

**Solution**

[latex]64\text{,}170[/latex]

**11. Identify the year in which the population will reach [latex]75\text{,}000[/latex].**

**Exercises: Algebraic Problems**

Instructions: For questions 12-17, consider this scenario: A town has an initial population of [latex]75\text{,}000[/latex]. It grows at a constant rate of [latex]2\text{,}500[/latex] per year for [latex]5[/latex] years.

**12. Find the linear function that models the town’s population [latex]P[/latex]as a function of the year, [latex]t[/latex], where[latex]t[/latex]is the number of years since the model began.**

**Solution**

[latex]P(t)=75,000+2500t[/latex]

**13. Find a reasonable domain and range for the function [latex]P[/latex].**

**14. If the function [latex]P[/latex] is graphed, find and interpret the [latex]x[/latex]– and [latex]y[/latex]-intercepts.**

**Solution**

[latex](–30,0)[/latex] Thirty years before the start of this model, the town had no citizens.

[latex](0,75\text{,}000)[/latex] Initially, the town had a population of [latex]75\text{,}000[/latex].

**15. If the function [latex]P[/latex] is graphed, find and interpret the slope of the function.**

**16. When will the population reach [latex]100\text{,}000[/latex]?**

**Solution**

Ten years after the model began

**17. What is the population in the year [latex]12[/latex] years from the onset of the model?**

**Exercises: Algebraic Problems**

Instructions: For questions 18-23, consider this scenario: The weight of a newborn is [latex]7.5[/latex] pounds. The baby gained one-half pound a month for its first year.

**18. Find the linear function that models the baby’s weight [latex]W[/latex]as a function of the age of the baby, in months, [latex]t[/latex].**

**Solution**

[latex]W(t)=0.5t+7.5[/latex]

**19. Find a reasonable domain and range for the function [latex]W[/latex].**

**20. If the function [latex]W[/latex] is graphed, find and interpret the [latex]x[/latex]– and [latex]y[/latex]-intercepts.**

**Solution**

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

**21. If the function [latex]W[/latex] is graphed, find and interpret the slope of the function.**

**22. When did the baby weight [latex]10.4[/latex] pounds?**

**Solution**

At age [latex]5.8[/latex] months

**23. What is the output when the input is [latex]6.2[/latex]?**

**Exercises: Algebraic Problems**

Instructions: For questions 24-29, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by [latex]205[/latex] each year from 2005 until 2010. In 2005, [latex]12\text{,}025[/latex] people were inflicted.

**24. Find the linear function that models the number of people inflicted with the common cold [latex]C[/latex] as a function of the year, [latex]t[/latex].**

**Solution**

[latex]C(t)=12\text{,}025-205t[/latex]

**25. Find a reasonable domain and range for the function [latex]C[/latex].**

**26. If the function [latex]C[/latex] is graphed, find and interpret the [latex]x[/latex]– and [latex]y[/latex]-intercepts.**

**Solution**

[latex](58.7,0)[/latex]: In roughly [latex]59[/latex] years, the number of people inflicted with the common cold would be [latex]0[/latex].

[latex](0,12\text{,}025)[/latex]: Initially there were [latex]12\text{,}025[/latex] people afflicted by the common cold.

**27. If the function [latex]C[/latex] is graphed, find and interpret the slope of the function.**

**28. When will the output reach [latex]0[/latex]?**

**Solution**

2063

**29. In what year will the number of people be [latex]9\text{,}700[/latex]?**

**Exercises: Graphical Problems**

Instructions: For questions 30-33, use the graph in Figure 3P.12.1, which shows the profit, [latex]y[/latex], in thousands of dollars, of a company in a given year, [latex]t[/latex], where [latex]t[/latex] represents the number of years since 1980.

**30. Find the linear function[latex]y[/latex], where[latex]y[/latex] depends on [latex]t[/latex], the number of years since 1980.**

**Solution**

[latex]y=-2t+180[/latex]

**31. Find and interpret the [latex]y[/latex]-intercept.**

**32. Find and interpret the [latex]x[/latex]-intercept.**

**Solution**

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

**33. Find and interpret the slope.**

**Exercises: ****Graphical Problems**

Instructions: For questions 34-37, use the graph in Figure 3P.12.2, which shows the profit, [latex]y[/latex],in thousands of dollars, of a company in a given year, [latex]t[/latex],where[latex]t[/latex]represents the number of years since 1980.

**34. Find the linear function[latex]y[/latex], where [latex]y[/latex] depends on [latex]t[/latex], the number of years since 1980.**

**Solution**

[latex]y=30t-300[/latex]

**35. Find and interpret the [latex]y[/latex]-intercept.**

**Solution**

[latex](0,-300)[/latex] In 1980, the company lost [latex]$300\text{,}000[/latex].

**36. Find and interpret the [latex]x[/latex]-intercept.**

**37. Find and interpret the slope.**

**Solution**

[latex]y=30t-300[/latex] of form [latex]y=mx+b,m=30[/latex]. For each year after 1980, the company’s profits increased [latex]$30\text{,}000[/latex] per year

**Exercises: Numeric Problems**

Instructions: For questions 38-40, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 3P.12.1. Assume that the house values are changing linearly.

Year | Mississippi | Hawaii |
---|---|---|

1950 | [latex]$25\text{,}200[/latex] | [latex]$74\text{,}400[/latex] |

2000 | [latex]$71\text{,}400[/latex] | [latex]$272\text{,}700[/latex] |

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

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

**Solution**

[latex]$80\text{,}640[/latex]

**40. 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.)**

**Exercises: Numeric**

Instructions: For questions 41-43, use the median home values in Indiana and Alabama (adjusted for inflation) shown in (Figure). Assume that the house values are changing linearly.

Year | Indiana | Alabama |
---|---|---|

1950 | $37,700 | $27,100 |

2000 | $94,300 | $85,100 |

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

**Solution**

Alabama

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

**43. 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**

2328

**Exercises: Real-World Applications**

**44. At noon, a barista notices that she has [latex]$20[/latex] in her tip jar. If she makes an average of [latex]$0.50[/latex] from each customer, how much will she have in her tip jar if she serves[latex]n[/latex]more customers during her shift?**

**Solution**

[latex]20+0.5n[/latex]

**45. A gym membership with two personal training sessions costs [latex]$125[/latex], while gym membership with five personal training sessions costs [latex]$260[/latex]. What is cost per session?**

**46. A clothing business finds there is a linear relationship between the number of shirts, [latex]n[/latex], it can sell and the price, [latex]p[/latex], it can charge per shirt. In particular, historical data shows that [latex]1\text{,}000[/latex] shirts can be sold at a price of [latex]$30[/latex], while [latex]3\text{,}000[/latex] shirts can be sold at a price of [latex]$22[/latex]. Find a linear equation in the form [latex]p(n)=mn+b[/latex] that gives the price [latex]p[/latex] they can charge for [latex]n[/latex] shirts.**

**Solution**

[latex]p(n)=-0.004n+34[/latex]

**47. A phone company charges for service according to the formula: [latex]C(n)=24+0.1n[/latex], where [latex]n[/latex] is the number of minutes talked, and [latex]C(n)[/latex] is the monthly charge, in dollars. Find and interpret the rate of change and initial value.**

**48. A farmer finds there is a linear relationship between the number of bean stalks, [latex]n[/latex], she plants and the yield, [latex]y[/latex], each plant produces. When she plants [latex]30[/latex] stalks, each plant yields [latex]30[/latex] oz of beans. When she plants [latex]34[/latex] stalks, each plant produces [latex]28[/latex] oz of beans. Find a linear relationships in the form [latex]y=mn+b[/latex] that gives the yield when [latex]n[/latex] stalks are planted.**

**Solution**

[latex]y=-0.5n+45[/latex]

**49. A city’s population in the year 1960 was [latex]287\text{,}500[/latex]. In 1989 the population was [latex]275\text{,}900[/latex]. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.**

**50. A town’s population has been growing linearly. In 2003, the population was [latex]45\text{,}000[/latex], and the population has been growing by [latex]1\text{,}700[/latex] people each year. Write an equation, [latex]P(t)[/latex], for the population [latex]t[/latex] years after 2003.**

**Solution**

[latex]P(t)=1700t+45,000[/latex]

**51. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: [latex]I(x)=1054x+23\text{,}286[/latex], where [latex]x[/latex] is the number of years after 1990. Which of the following interprets the slope in the context of the problem?**

**a. As of 1990, average annual income was [latex]$23\text{,}286[/latex].**

**b. In the ten-year period from 1990–1999, average annual income increased by a total of [latex]$1\text{,}054[/latex].**

**c. Each year in the decade of the 1990s, average annual income increased by [latex]$1\text{,}054[/latex].**

**d. Average annual income rose to a level of [latex]$23\text{,}286[/latex] by the end of 1999.**

**52. When temperature is [latex]0[/latex] degrees Celsius, the Fahrenheit temperature is [latex]32[/latex]. When the Celsius temperature is [latex]100[/latex], the corresponding Fahrenheit temperature is [latex]212[/latex]. Express the Fahrenheit temperature as a linear function of [latex]C[/latex], the Celsius temperature, [latex]F(C)[/latex].**

**a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.**

**b. Find and interpret [latex]F(28)[/latex].**

**c. Find and interpret [latex]F(–40)[/latex].**

**Solution**

a. [latex]\text{Rate of change =}\frac{\Delta F}{\Delta C}=\frac{212-32}{100-0}=1.8\text{ degrees F for one degree change in C}[/latex]

b. [latex]F(28)=1.8(28)+32=82.4\text{ degrees F is 28 degrees C}[/latex]

c. [latex]F(-40)=1.8(-40)+32=-40\text{ degrees F is -40 degrees C}[/latex]

**53. In 2004, a school population was [latex]1001[/latex]. By 2008 the population had grown to [latex]1697[/latex]. Assume the population is changing linearly.**

**a. How much did the population grow between the year 2004 and 2008?**

**b. How long did it take the population to grow from [latex]1001[/latex] students to [latex]1697[/latex] students?**

**c. What is the average population growth per year?**

**d. What was the population in the year 2000?**

**e. Find an equation for the population, [latex]P[/latex], of the school [latex]t[/latex] years after 2000.**

**f. Using your equation, predict the population of the school in 2011.**

**54. In 2003, a town’s population was [latex]1431[/latex]. By 2007 the population had grown to [latex]2134[/latex]. Assume the population is changing linearly.**

**a. How much did the population grow between the year 2003 and 2007?**

**b. How long did it take the population to grow from [latex]1431[/latex] people to [latex]2134[/latex] people?**

**c. What is the average population growth per year?**

**d. What was the population in the year 2000?**

**e. Find an equation for the population, [latex]P[/latex], of the town [latex]t[/latex] years after 2000.**

**f. Using your equation, predict the population of the town in 2014.**

**Solution**

a. [latex]2134-1431=703[/latex] people

b. [latex]2007-2003=4[/latex] years

c. Average rate of growth [latex]=\frac{703}{4}=175.75[/latex] people per year. So, using [latex]y=mx+b[/latex], we have [latex]y=175.75x+1431[/latex].

d. The year 2000 corresponds to [latex]t=-3[/latex].

So, [latex]y=175.75(-3)+1431=903.75[/latex] or roughly [latex]904[/latex] people in year 2000

e. If the year 2000 corresponds to [latex]t=0[/latex], then we have ordered pair [latex](0,903.75)[/latex].

[latex]y=175.75x+903.75[/latex] corresponds to [latex]P(t)=175.75t+903.75[/latex].

f. The year 2014 corresponds to [latex]t=14[/latex]. Therefore, [latex]P(14)=175.75(14)+903.75=3364.25[/latex].

So, a population of 3364.

**55. 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 on the phone. If a customer uses [latex]410[/latex] minutes, the monthly cost will be [latex]$71.50[/latex]. If the customer uses [latex]720[/latex] minutes, the monthly cost will be [latex]$118[/latex].**

**a. Find a linear equation for the monthly cost of the cell plan as a function of [latex]x[/latex], the number of monthly minutes used.**

**b. Interpret the slope and [latex]y[/latex]-intercept of the equation.**

**c. Use your equation to find the total monthly cost if [latex]687[/latex] minutes are used.**

**56. A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of [latex]$10[/latex] and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses [latex]20[/latex] MB, the monthly cost will be [latex]$11.20[/latex]. If the customer uses [latex]130[/latex] MB, the monthly cost will be [latex]$17.80[/latex].**

**a. Find a linear equation for the monthly cost of the data plan as a function of [latex]x[/latex], the number of MB used.**

**b. Interpret the slope and [latex]y[/latex]-intercept of the equation.**

**c. Use your equation to find the total monthly cost if [latex]250[/latex] MB are used.**

**Solution**

a.

[latex]\begin{array}{l}\text{Ordered pairs are }(20,11.20)\text{ and }(130,17.80)\\\begin{array}{rcl}m&=&\frac{17.80-11.20}{130-20}=0.06\text{ and }b\text{ is at the point }(0,10)\\y&=&mx+b\\y&=& 0.06x+10\\\text{or }C(x)&=&0.06x+10 \end{array}\end{array}[/latex]

b.

[latex]0.06[/latex] – For every MB, the client is charged [latex]6[/latex] cents.

[latex](0,10)[/latex] – If no usage occurs, the client is charged [latex]$10[/latex].

c.

[latex]\begin{array}{rcl}C(250)&=& 0.06(250)+10\\&=&$25\end{array}[/latex]

**57. In 1991, the moose population in a park was measured to be [latex]4\text{,}360[/latex]. By 1999, the population was measured again to be [latex]5\text{,}880[/latex]. Assume the population continues to change linearly.**

**a. Find a formula for the moose population, [latex]P[/latex] since 1990.**

**b. What does your model predict the moose population to be in 2003?**

**58. In 2003, the owl population in a park was measured to be [latex]340[/latex]. By 2007, the population was measured again to be [latex]285[/latex]. The population changes linearly. Let the input be years since 1990.**

**a. Find a formula for the owl population, [latex]P[/latex]. Let the input be years since 2003.**

**b. What does your model predict the owl population to be in 2012?**

**Solution**

a.

[latex]\begin{array}{l}\text{Ordered pairs are }(0,340)\text{ and }(4,285)\\\begin{array}{rcl}m&=&\frac{285-340}{4-0}=-13.75\text{ and }b\text{ is at the point }(0,340)\\y&=&mx+b\\y&=&-13.75x+340 \\ \text{or }P(t)&=&-13.75t+340\hfill \end{array}\end{array}[/latex]

b.

[latex]\begin{array}{l}\text{The year 2012 corresponds to }t=9\\\begin{array}{rcl}P(9)&=&-13.75(9)+340\\&=&216.25\text{ or }216\text{ owls}\end{array}\end{array}[/latex]

**59. The Federal Helium Reserve held about [latex]16[/latex] billion cubic feet of helium in 2010 and is being depleted by about [latex]2.1[/latex] billion cubic feet each year.**

**a. Give a linear equation for the remaining federal helium reserves, [latex]R[/latex], in terms of [latex]t[/latex], the number of years since 2010.**

**b. In 2015, what will the helium reserves be?**

**c. If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?**

**60. Suppose the world’s oil reserves in 2014 are [latex]1\text{,}820[/latex] billion barrels. If, on average, the total reserves are decreasing by [latex]25[/latex] billion barrels of oil each year:**

**a. Give a linear equation for the remaining oil reserves, [latex]R[/latex], in terms of [latex]t[/latex], the number of years since now.**

**b. Seven years from now, what will the oil reserves be?**

**c. If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?**

**Solution**

a.

[latex]\begin{array}{l}\text{The year 2014 corresponds to }t=0.\\ \text{We have }m=-25\text{ and }b\text{ at the point }(0,1820)\\ \begin{array}{rcl}y&=&mx+b\\y&=&-25x+1820\\\text{or }R(t)&=&-25t+1820\end{array}\end{array}[/latex]

b.

[latex]\begin{array}{rcl}R(7)&=&-25(7)+1820\\ &=& 645\text{ billion cubic feet}\end{array}[/latex]

c.

[latex]\begin{array}{rcl}0&=&-25t+1820\\-1820&=&-25t\\72.8&=&t\;⇒\;2014+72.8=2086.8\text{. So, in the year 2086}\end{array}[/latex]

**61. You are choosing between two different prepaid cell phone plans. The first plan charges a rate of [latex]26[/latex] cents per minute. The second plan charges a monthly fee of [latex]$19.95[/latex]**

*plus*[latex]11[/latex] cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?**62. You are choosing between two different window washing companies. The first charges [latex]$5[/latex] per window. The second charges a base fee of [latex]$40[/latex] plus [latex]$3[/latex] per window. How many windows would you need to have for the second company to be preferable?**

**Solution**

[latex]\begin{array}{l}\text{Plan 1: }y=5x\text{ where }x\text{ is number of windows}\\ \text{Plan 2: }y=3x+40\text{ where }x\text{ is number of windows}\\ \begin{array}{rcl}3x+40&\le&5x\\40&\le&2x\\20&\le&x\end{array}\\ \text{So, more than 20 windows}\end{array}[/latex]

**63. When hired at a new job selling jewelry, you are given two pay options:**

**Option A: Base salary of [latex]$17\text{,}000[/latex] a year with a commission of [latex]12\%[/latex] of your sales**

**Option B: Base salary of [latex]$20\text{,}000[/latex] a year with a commission of [latex]5\%[/latex] of your sales**

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

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

**Option A: Base salary of [latex]$14\text{,}000[/latex] a year with a commission of [latex]10\%[/latex] of your sales**

**Option B: Base salary of [latex]$19\text{,}000[/latex] a year with a commission of [latex]4\%[/latex] of your sales**

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

**Solution**

[latex]\begin{array}{l}\text{Option A: }y=0.10x+14\text{,}000\text{ where }x\text{ is dollars of sales.}\\\text{Option B: }y=0.04x+19\text{,}000\text{ where }x\text{ is dollars of sales.}\\ \begin{array}{rcl}0.10x+14\text{,}000&\ge&0.04x+19\text{,}000\\0.06x+14\text{,}000&\ge&19\text{,}000\\0.06x&\ge&5\text{,}000\\x&\ge&83\text{,}333.33\end{array}\\\text{So, more than }$83\text{,}333.33\text{ in sales.}\end{array}[/latex]

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

**Option A: Base salary of [latex]$20\text{,}000[/latex] a year with a commission of [latex]12\%[/latex] of your sales**

**Option B: Base salary of [latex]$26\text{,}000[/latex] a year with a commission of [latex]3\%[/latex] of your sales**

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

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

**Option A: Base salary of [latex]$10\text{,}000[/latex] a year with a commission of [latex]9\%[/latex] of your sales**

**Option B: Base salary of [latex]$20\text{,}000[/latex] a year with a commission of [latex]4\%[/latex] of your sales**

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

**Solution**

[latex]\begin{array}{l}\text{Option A: }y=0.09x+10\text{,}000\text{ where }x\text{ is dollars of sales.}\\ \text{Option B: }y=0.04x+20\text{,}000\text{ where }x\text{ is dollars of sales.}\\ \begin{array}{rcl}0.09x+10\text{,}000&\ge&0.04x+20\text{,}000\\0.05x+10\text{,}000&\ge&20\text{,}000\\0.05x&\ge&10\text{,}000\\x&\ge& 200\text{,}000\end{array}\\ \text{So, more than }$200\text{,}000\text{ in sales.}\end{array}[/latex]