1.4 Applications
Section 1.4 Applications
In this section, we apply linear functions to model real-world situations. Whether you're calculating business costs, converting temperatures, or predicting population growth, linear models give us powerful tools to make predictions and understand relationships between quantities.
When approaching application problems, read carefully and identify which values correspond to the independent variable ($x$) and which correspond to the dependent variable ($y$).
Why bother modeling the real world with linear functions? In business, these models help you predict costs and set prices. In science, they help you understand how variables like temperature and pressure relate. In everyday life, they can help you compare cell phone plans or calculate gas mileage. When we find a linear relationship, we gain the ability to predict future outcomes based on current patterns—without having to test every possible scenario.
Taxi Service Costs
A taxi service charges $\$0.50$ per mile plus a $\$5$ flat fee. What will be the cost of traveling 20 miles? What will be the cost of traveling $x$ miles?
Example 1.4.1 Solution
First, we identify our variables. Let $\mathbf{x} =$ the distance traveled in miles, and let $\mathbf{y} =$ the total cost in dollars.
To find the cost of traveling 20 miles, we substitute $x = 20$ into our relationship: \[y = 0.50(20) + 5 = 10 + 5 = 15\] The cost for 20 miles is $\$15$.
To find the cost of traveling $x$ miles, we write a general cost function: \[y = 0.50x + 5\]
In this equation, $\$0.50$ per mile represents the variable cost (it varies with distance), and the $\$5$ flat fee represents the fixed cost (it remains constant regardless of distance). Notice how this matches the slope-intercept form $y = mx + b$: the variable cost corresponds to the slope ($m = 0.50$), and the fixed cost corresponds to the y-intercept ($b = 5$).
In business and economics, distinguishing between variable cost and fixed cost is essential for break-even analysis. Fixed costs (like rent or equipment leases) remain the same whether you produce 1 unit or 1,000 units. Variable costs (like raw materials or fuel) scale directly with production. Knowing these components helps business owners calculate exactly how many units they need to sell to start making a profit.
Think of a linear cost equation like a smartphone data plan. You pay a fixed monthly fee (the y-intercept) plus a charge for every gigabyte of data you use (the slope). Understanding which part is fixed and which is variable helps you compare different plans and find the best deal for your specific usage habits.
Manufacturing Costs
The variable cost to manufacture a product is $\$10$ per item, and the fixed cost is $\$2500$. If $x$ represents the number of items manufactured and $y$ represents the total cost, write the cost function.
Example 1.4.2 Solution
We are given the components of a linear cost function directly:
- The variable cost of $\$10$ per item tells us the rate of change, so our slope $m = 10$.
- The fixed cost represents the initial value when production is zero ($x = 0$), so our y-intercept $b = 2500$.
Substituting these into the slope-intercept form $y = mx + b$, we get the cost equation: \[y = 10x + 2500\]
Finding the Cost Equation from Data
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 function to predict the cost of 100 items.
Example 1.4.3 Solution
We define our variables: let $\mathbf{x} =$ the number of items manufactured, and let $\mathbf{y} =$ the total cost.
This problem gives us two data points: $(25, 750)$ and $(50, 1000)$. Finding the cost equation is equivalent to finding the equation of a line passing through these two points.
First, we calculate the slope $m$: \[m = \frac{1000 - 750}{50 - 25} = \frac{250}{25} = 10\]
This gives us the partial equation $y = 10x + b$. To find $b$, we substitute either point into the equation. Using $(25, 750)$: \[750 = 10(25) + b\] \[750 = 250 + b\] \[b = 500\]
Therefore, the cost equation is: \[y = 10x + 500\]
To predict the cost of 100 items, we substitute $x = 100$: \[y = 10(100) + 500 = 1000 + 500 = 1500\]
It costs $\$1500$ to manufacture 100 items.
Temperature Conversion
The freezing temperature of water is $0^\circ$C (Celsius) and $32^\circ$F (Fahrenheit). The boiling temperature of water is $100^\circ$C and $212^\circ$F. Write a conversion equation from Celsius to Fahrenheit, and use this equation to convert $30^\circ$C into Fahrenheit.
Example 1.4.4 Solution
We organize the given information as coordinate pairs $(C, F)$:
| Celsius ($C$) | Fahrenheit ($F$) |
|---|---|
| 0 | 32 |
| 100 | 212 |
We are looking for a linear equation of the form $F = mC + b$, where $C$ represents the temperature in Celsius and $F$ represents the temperature in Fahrenheit.
First, we find the slope $m$: \[m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}\]
So our equation is $F = \frac{9}{5}C + b$. To find $b$, we substitute the point $(0, 32)$: \[32 = \frac{9}{5}(0) + b\] \[b = 32\]
Therefore, the conversion equation is: \[F = \frac{9}{5}C + 32\]
To convert $30^\circ$C into Fahrenheit, we substitute $C = 30$: \[F = \frac{9}{5}(30) + 32 = 54 + 32 = 86\]
Thus, $30^\circ$C is equivalent to $86^\circ$F.
Population Modeling
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.
a. Write the linear function that gives a relationship between the time and the population.
b. Assuming the population continues to grow linearly in the future, use this equation to predict the population of Canada in the year 2025.
Example 1.4.5 Solution
To simplify the calculations, we use 1980 as the base year (year 0). This means:
- 1980 corresponds to $x = 0$
- 2010 corresponds to $x = 30$ (30 years after 1980)
- 2025 will correspond to $x = 45$ (45 years after 1980)
Our data points are $(0, 24.5)$ and $(30, 34)$, where $y$ is measured in millions:
| Year | $x$ (years after 1980) | Population (millions) |
|---|---|---|
| 1980 | 0 | 24.5 |
| 2010 | 30 | 34 |
Part a: We find the slope using the two points: \[m = \frac{34 - 24.5}{30 - 0} = \frac{9.5}{30} \approx 0.32\]
The y-intercept occurs at $x = 0$, which gives us $b = 24.5$.
Therefore, the population function is: \[y = 0.32x + 24.5\]
Part b: To predict the population in 2025, we calculate $x = 2025 - 1980 = 45$: \[y = 0.32(45) + 24.5\] \[y = 14.4 + 24.5 = 38.9\]
We predict the population of Canada will be 38.9 million people in the year 2025.
Important Note: This prediction assumes the linear trend continues indefinitely. If population growth patterns change (due to policy shifts, economic changes, or other factors), this model may become inaccurate. Always verify whether your assumptions remain valid when extrapolating with linear models.
Problem Set 1.4
In the following application problems, assume a linear relationship holds.
1) The variable cost to manufacture a product is $\$25$ per item, and the fixed costs are $\$1200$. If $x$ is the number of items manufactured and $y$ is the cost, write the cost function.
Problem 1 Solution
We identify the components of a linear cost function $y = mx + b$:
Step 1: Identify the slope (variable cost).
The variable cost is $\$25$ per item, so $m = 25$.
Step 2: Identify the y-intercept (fixed cost).
The fixed cost is $\$1200$, so $b = 1200$.
Step 3: Write the cost function by substituting into $y = mx + b$: \[y = 25x + 1200\]
Answer: $y = 25x + 1200$
Verification: If $x = 0$ items: $y = 25(0) + 1200 = 1200$ (matches the fixed cost). If $x = 10$ items: $y = 25(10) + 1200 = 250 + 1200 = 1450$, which is 10 items at $\$25$ each plus the $\$1200$ fixed cost. ✓
2) It costs $\$90$ to rent a car driven 100 miles and $\$140$ for one driven 200 miles. If $x$ is the number of miles driven and $y$ the total cost of the rental, write the cost function.
Problem 2 Solution
We are given two data points: $(100, 90)$ and $(200, 140)$.
Step 1: Calculate the slope $m$: \[m = \frac{140 - 90}{200 - 100} = \frac{50}{100} = 0.50\]
This means the rental costs $\$0.50$ per mile driven.
Step 2: Find the y-intercept $b$ using the point $(100, 90)$: \[90 = 0.50(100) + b\] \[90 = 50 + b\] \[b = 40\]
The $\$40$ represents the base rental fee before any miles are driven.
Step 3: Write the cost function: \[y = 0.50x + 40\]
Answer: $y = 0.50x + 40$ (or equivalently $y = \frac{1}{2}x + 40$)
Verification: For $x = 100$: $y = 0.50(100) + 40 = 50 + 40 = 90$ ✓. For $x = 200$: $y = 0.50(200) + 40 = 100 + 40 = 140$ ✓
3) The variable cost to manufacture an item is $\$20$ per item, and it costs a total of $\$750$ to produce 20 items. If $x$ represents the number of items manufactured and $y$ is the cost, write the cost function.
Problem 3 Solution
We know the slope (variable cost) and one data point.
Step 1: Identify the slope.
The variable cost is $\$20$ per item, so $m = 20$.
Step 2: Use the data point $(20, 750)$ to find $b$: \[750 = 20(20) + b\] \[750 = 400 + b\] \[b = 350\]
The fixed cost is $\$350$.
Step 3: Write the cost function: \[y = 20x + 350\]
Answer: $y = 20x + 350$
Verification: For $x = 20$: $y = 20(20) + 350 = 400 + 350 = 750$ ✓
4) To manufacture 30 items, it costs $\$2700$, and to manufacture 50 items, it costs $\$3200$. If $x$ represents the number of items manufactured and $y$ the cost, write the cost function.
Problem 4 Solution
We are given two data points: $(30, 2700)$ and $(50, 3200)$.
Step 1: Calculate the slope $m$: \[m = \frac{3200 - 2700}{50 - 30} = \frac{500}{20} = 25\]
The variable cost is $\$25$ per item.
Step 2: Find the y-intercept $b$ using the point $(30, 2700)$: \[2700 = 25(30) + b\] \[2700 = 750 + b\] \[b = 1950\]
The fixed cost is $\$1950$.
Step 3: Write the cost function: \[y = 25x + 1950\]
Answer: $y = 25x + 1950$
Verification: For $x = 30$: $y = 25(30) + 1950 = 750 + 1950 = 2700$ ✓. For $x = 50$: $y = 25(50) + 1950 = 1250 + 1950 = 3200$ ✓
5) To manufacture 100 items, it costs $\$32,000$, and to manufacture 200 items, it costs $\$40,000$. If $x$ is the number of items manufactured and $y$ is the cost, write the cost function.
Problem 5 Solution
We are given two data points: $(100,\ 32000)$ and $(200,\ 40000)$.
Step 1: Calculate the slope $m$: \[m = \frac{40000 - 32000}{200 - 100} = \frac{8000}{100} = 80\]
The variable cost is $\$80$ per item.
Step 2: Find the y-intercept $b$ using the point $(100, 32000)$: \[32000 = 80(100) + b\] \[32000 = 8000 + b\] \[b = 24000\]
The fixed cost is $\$24{,}000$.
Step 3: Write the cost function: \[y = 80x + 24000\]
Answer: $y = 80x + 24{,}000$
Verification: For $x = 100$: $y = 80(100) + 24000 = 8000 + 24000 = 32000$ ✓. For $x = 200$: $y = 80(200) + 24000 = 16000 + 24000 = 40000$ ✓
6) It costs $\$1900$ to manufacture 60 items, and the fixed costs are $\$700$. If $x$ represents the number of items manufactured and $y$ the cost, write the cost function.
Problem 6 Solution
We know the fixed cost (y-intercept) and one data point.
Step 1: Identify the y-intercept.
The fixed cost is $\$700$, so $b = 700$.
Step 2: Use the data point $(60, 1900)$ to find the slope $m$: \[1900 = m(60) + 700\] \[1200 = 60m\] \[m = \frac{1200}{60} = 20\]
The variable cost is $\$20$ per item.
Step 3: Write the cost function: \[y = 20x + 700\]
Answer: $y = 20x + 700$
Verification: For $x = 60$: $y = 20(60) + 700 = 1200 + 700 = 1900$ ✓. For $x = 0$: $y = 700$ (matches the given fixed cost) ✓
7) A person who weighs 150 pounds has 60 pounds of muscles; a person that weighs 180 pounds has 72 pounds of muscles. If $x$ represents body weight and $y$ is muscle weight, write an equation describing their relationship. Use this relationship to determine the muscle weight of a person that weighs 170 pounds.
Problem 7 Solution
We are given two data points: $(150, 60)$ and $(180, 72)$.
Step 1: Calculate the slope $m$: \[m = \frac{72 - 60}{180 - 150} = \frac{12}{30} = 0.4\]
This means for every additional pound of body weight, muscle weight increases by $0.4$ pounds.
Step 2: Find the y-intercept $b$ using the point $(150, 60)$: \[60 = 0.4(150) + b\] \[60 = 60 + b\] \[b = 0\]
Step 3: Write the equation: \[y = 0.4x\]
This tells us muscle weight is always 40% of body weight.
Step 4: Predict muscle weight for a 170-pound person: \[y = 0.4(170) = 68\]
Answer: The equation is $y = 0.4x$. A person weighing 170 pounds has 68 pounds of muscle.
Verification: For $x = 150$: $y = 0.4(150) = 60$ ✓. For $x = 180$: $y = 0.4(180) = 72$ ✓
8) A spring on a door stretches 6 inches if a force of 30 pounds is applied. It stretches 10 inches if a 50 pound force is applied. If $x$ represents the number of inches stretched, and $y$ is the force, write a linear equation describing the relationship. Use it to determine the amount of force required to stretch the spring 12 inches.
Problem 8 Solution
We are given two data points where $x =$ inches stretched and $y =$ force in pounds: $(6, 30)$ and $(10, 50)$.
Step 1: Calculate the slope $m$: \[m = \frac{50 - 30}{10 - 6} = \frac{20}{4} = 5\]
This means each additional inch of stretch requires 5 additional pounds of force.
Step 2: Find the y-intercept $b$ using the point $(6, 30)$: \[30 = 5(6) + b\] \[30 = 30 + b\] \[b = 0\]
Step 3: Write the equation: \[y = 5x\]
This is consistent with Hooke's Law — force is directly proportional to displacement.
Step 4: Find the force required to stretch the spring 12 inches: \[y = 5(12) = 60\]
Answer: The equation is $y = 5x$. A force of 60 pounds is required to stretch the spring 12 inches.
Verification: For $x = 6$: $y = 5(6) = 30$ ✓. For $x = 10$: $y = 5(10) = 50$ ✓
9) A male college student who is 64 inches tall weighs 110 pounds. Another student who is 74 inches tall weighs 180 pounds. Assuming the relationship between male students' heights ($x$), and weights ($y$) is linear, write a function to express weights in terms of heights, and use this function to predict the weight of a student who is 68 inches tall.
Problem 9 Solution
We are given two data points: $(64, 110)$ and $(74, 180)$.
Step 1: Calculate the slope $m$: \[m = \frac{180 - 110}{74 - 64} = \frac{70}{10} = 7\]
This means each additional inch of height corresponds to 7 additional pounds of weight.
Step 2: Find the y-intercept $b$ using the point $(64, 110)$: \[110 = 7(64) + b\] \[110 = 448 + b\] \[b = 110 - 448 = -338\]
Step 3: Write the weight function: \[y = 7x - 338\]
Step 4: Predict the weight for a student who is 68 inches tall: \[y = 7(68) - 338 = 476 - 338 = 138\]
Answer: The function is $y = 7x - 338$. A student who is 68 inches tall is predicted to weigh 138 pounds.
Verification: For $x = 64$: $y = 7(64) - 338 = 448 - 338 = 110$ ✓. For $x = 74$: $y = 7(74) - 338 = 518 - 338 = 180$ ✓
10) EZ Clean company has determined that if it spends $\$30,000$ on advertising, it can hope to sell 12,000 of its Minivacs a year, but if it spends $\$50,000$, it can sell 16,000. Write an equation that gives a relationship between the number of dollars spent on advertising ($x$) and the number of minivacs sold ($y$).
Problem 10 Solution
We are given two data points: $(30000, 12000)$ and $(50000, 16000)$.
Step 1: Calculate the slope $m$: \[m = \frac{16000 - 12000}{50000 - 30000} = \frac{4000}{20000} = 0.2\]
This means for every additional dollar spent on advertising, 0.2 additional Minivacs are sold (or equivalently, every $\$5$ in advertising yields 1 additional sale).
Step 2: Find the y-intercept $b$ using the point $(30000, 12000)$: \[12000 = 0.2(30000) + b\] \[12000 = 6000 + b\] \[b = 6000\]
The company can expect to sell 6,000 Minivacs even without advertising.
Step 3: Write the equation: \[y = 0.2x + 6000\]
Answer: $y = 0.2x + 6000$
Verification: For $x = 30000$: $y = 0.2(30000) + 6000 = 6000 + 6000 = 12000$ ✓. For $x = 50000$: $y = 0.2(50000) + 6000 = 10000 + 6000 = 16000$ ✓
11) The freezing temperatures for water for Celsius and Fahrenheit scales are $0^\circ$C and $32^\circ$F. The boiling temperatures for water are $100^\circ$C and $212^\circ$F. Let $C$ denote the temperature in Celsius and $F$ in Fahrenheit. Write the conversion function from Celsius to Fahrenheit. Use the function to convert $25^\circ$C into $^\circ$F.
Problem 11 Solution
We are given two data points $(C, F)$: $(0, 32)$ and $(100, 212)$.
Step 1: Calculate the slope $m$: \[m = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}\]
Each degree Celsius corresponds to $\frac{9}{5}$ degrees Fahrenheit.
Step 2: Find the y-intercept $b$.
Since the point $(0, 32)$ is on the line, the y-intercept is $b = 32$.
Step 3: Write the conversion function: \[F = \frac{9}{5}C + 32\]
Step 4: Convert $25^\circ$C to Fahrenheit: \[F = \frac{9}{5}(25) + 32 = \frac{225}{5} + 32 = 45 + 32 = 77\]
Answer: The conversion function is $F = \frac{9}{5}C + 32$. Thus $25^\circ$C $= \mathbf{77^\circ}$F.
Verification: For $C = 0$: $F = \frac{9}{5}(0) + 32 = 32$ ✓ (freezing point). For $C = 100$: $F = \frac{9}{5}(100) + 32 = 180 + 32 = 212$ ✓ (boiling point).
12) By reversing the coordinates in the previous problem, find a conversion function that converts Fahrenheit into Celsius, and use this conversion function to convert $72^\circ$F into an equivalent Celsius measure.
Problem 12 Solution
Reversing the coordinates from Problem 11, we treat $F$ as input and $C$ as output. Our data points $(F, C)$ are: $(32, 0)$ and $(212, 100)$.
Step 1: Calculate the slope $m$: \[m = \frac{100 - 0}{212 - 32} = \frac{100}{180} = \frac{5}{9}\]
Step 2: Find the y-intercept $b$ using the point $(32, 0)$: \[0 = \frac{5}{9}(32) + b\] \[0 = \frac{160}{9} + b\] \[b = -\frac{160}{9}\]
Step 3: Write the conversion function: \[C = \frac{5}{9}F - \frac{160}{9} = \frac{5}{9}(F - 32)\]
Step 4: Convert $72^\circ$F to Celsius: \[C = \frac{5}{9}(72 - 32) = \frac{5}{9}(40) = \frac{200}{9} \approx 22.2\]
Answer: The conversion function is $C = \frac{5}{9}(F - 32)$. Thus $72^\circ$F $\approx \mathbf{22.2^\circ}$C.
Verification: For $F = 32$: $C = \frac{5}{9}(32 - 32) = 0$ ✓ (freezing). For $F = 212$: $C = \frac{5}{9}(212 - 32) = \frac{5}{9}(180) = 100$ ✓ (boiling).
13) California's population was 29.8 million in the year 1990, and 37.3 million in 2010. Assume that the population trend was and continues to be linear, write the population function. Use this function to predict the population in 2025. Hint: Use 1990 as the base year (year 0); then 2010 and 2025 are years 20, and 35, respectively.
Problem 13 Solution
Let $x =$ years after 1990 and $y =$ population in millions. Our data points are: $(0, 29.8)$ and $(20, 37.3)$.
Step 1: Calculate the slope $m$: \[m = \frac{37.3 - 29.8}{20 - 0} = \frac{7.5}{20} = 0.375\]
The population grows by 0.375 million (375,000) people per year.
Step 2: Identify the y-intercept.
Since $(0, 29.8)$ is a data point, $b = 29.8$.
Step 3: Write the population function: \[y = 0.375x + 29.8\]
Step 4: Predict the population in 2025 ($x = 2025 - 1990 = 35$): \[y = 0.375(35) + 29.8 = 13.125 + 29.8 = 42.925\]
Answer: The population function is $y = 0.375x + 29.8$ (in millions). The predicted population of California in 2025 is approximately 42.925 million people.
Verification: For $x = 0$ (1990): $y = 0.375(0) + 29.8 = 29.8$ ✓. For $x = 20$ (2010): $y = 0.375(20) + 29.8 = 7.5 + 29.8 = 37.3$ ✓
14) Use the population function for California in the previous problem to find the year in which the population will be 40 million people.
Problem 14 Solution
From Problem 13, the population function is $y = 0.375x + 29.8$, where $x$ is years after 1990.
Step 1: Set $y = 40$ and solve for $x$: \[40 = 0.375x + 29.8\]
Step 2: Subtract 29.8 from both sides: \[40 - 29.8 = 0.375x\] \[10.2 = 0.375x\]
Step 3: Divide both sides by 0.375: \[x = \frac{10.2}{0.375} = 27.2\]
Step 4: Convert back to a calendar year: \[1990 + 27.2 = 2017.2\]
Answer: The population of California will reach 40 million during the year 2017.
Verification: $y = 0.375(27.2) + 29.8 = 10.2 + 29.8 = 40.0$ ✓
15) A college's enrollment was 13,200 students in the year 2000, and 15,000 students in 2015. Enrollment has followed a linear pattern. Write the function that models enrollment as a function of time. Use the function to find the college's enrollment in the year 2010. Hint: Use year 2000 as the base year.
Problem 15 Solution
Let $x =$ years after 2000 and $y =$ enrollment. Our data points are: $(0, 13200)$ and $(15, 15000)$.
Step 1: Calculate the slope $m$: \[m = \frac{15000 - 13200}{15 - 0} = \frac{1800}{15} = 120\]
Enrollment increases by 120 students per year.
Step 2: Identify the y-intercept.
Since $(0, 13200)$ is a data point, $b = 13200$.
Step 3: Write the enrollment function: \[y = 120x + 13200\]
Step 4: Find enrollment in 2010 ($x = 2010 - 2000 = 10$): \[y = 120(10) + 13200 = 1200 + 13200 = 14400\]
Answer: The enrollment function is $y = 120x + 13{,}200$. The college's enrollment in 2010 was 14,400 students.
Verification: For $x = 0$ (2000): $y = 120(0) + 13200 = 13200$ ✓. For $x = 15$ (2015): $y = 120(15) + 13200 = 1800 + 13200 = 15000$ ✓
16) If the college's enrollment continues to follow this pattern, in what year will the college have 16,000 students enrolled.
Problem 16 Solution
From Problem 15, the enrollment function is $y = 120x + 13200$, where $x$ is years after 2000.
Step 1: Set $y = 16000$ and solve for $x$: \[16000 = 120x + 13200\]
Step 2: Subtract 13200 from both sides: \[16000 - 13200 = 120x\] \[2800 = 120x\]
Step 3: Divide both sides by 120: \[x = \frac{2800}{120} = 23.\overline{3}\]
Step 4: Convert back to a calendar year: \[2000 + 23.33 \approx 2023.33\]
Answer: The college's enrollment will reach 16,000 students during the year 2023.
Verification: $y = 120(23.\overline{3}) + 13200 = 2800 + 13200 = 16000$ ✓
17) The cost of electricity in residential homes is a linear function of the amount of energy used. In Grove City, a home using 250 kilowatt hours (kwh) of electricity per month pays $\$55$. A home using 600 kwh per month pays $\$118$. Write the cost of electricity as a function of the amount used. Use the function to find the cost for a home using 400 kwh of electricity per month.
Problem 17 Solution
We are given two data points where $x =$ kwh used and $y =$ monthly cost: $(250, 55)$ and $(600, 118)$.
Step 1: Calculate the slope $m$: \[m = \frac{118 - 55}{600 - 250} = \frac{63}{350} = 0.18\]
The cost is $\$0.18$ per kilowatt hour.
Step 2: Find the y-intercept $b$ using the point $(250, 55)$: \[55 = 0.18(250) + b\] \[55 = 45 + b\] \[b = 10\]
The base monthly charge (before any electricity use) is $\$10$.
Step 3: Write the cost function: \[y = 0.18x + 10\]
Step 4: Find the cost for 400 kwh: \[y = 0.18(400) + 10 = 72 + 10 = 82\]
Answer: The cost function is $y = 0.18x + 10$. The monthly cost for 400 kwh is $\$82$.
Verification: For $x = 250$: $y = 0.18(250) + 10 = 45 + 10 = 55$ ✓. For $x = 600$: $y = 0.18(600) + 10 = 108 + 10 = 118$ ✓
18) Find the level of electricity use that would correspond to a monthly cost of $\$100$.
Problem 18 Solution
From Problem 17, the electricity cost function is $y = 0.18x + 10$.
Step 1: Set $y = 100$ and solve for $x$: \[100 = 0.18x + 10\]
Step 2: Subtract 10 from both sides: \[90 = 0.18x\]
Step 3: Divide both sides by 0.18: \[x = \frac{90}{0.18} = 500\]
Answer: A monthly cost of $\$100$ corresponds to 500 kwh of electricity use.
Verification: $y = 0.18(500) + 10 = 90 + 10 = 100$ ✓
19) At ABC Co., sales revenue is $\$170,000$ when it spends $\$5,000$ on advertising. Sales revenue is $\$254,000$ when $\$12,000$ is spent on advertising.
a) Find a linear function for $y =$ amount of sales revenue as a function of $x =$ amount spent on advertising.
b) Find revenue if $\$10,000$ is spent on advertising.
c) Find the amount that should be spent on advertising to achieve $\$200,000$ in revenue.
Problem 19 Solution
We are given two data points: $(5000, 170000)$ and $(12000, 254000)$.
Part a)
Step 1: Calculate the slope $m$: \[m = \frac{254000 - 170000}{12000 - 5000} = \frac{84000}{7000} = 12\]
For every additional dollar spent on advertising, revenue increases by $\$12$.
Step 2: Find the y-intercept $b$ using the point $(5000, 170000)$: \[170000 = 12(5000) + b\] \[170000 = 60000 + b\] \[b = 110000\]
Step 3: Write the revenue function: \[y = 12x + 110{,}000\]
Part b)
Substitute $x = 10000$: \[y = 12(10000) + 110000 = 120000 + 110000 = 230000\]
Revenue would be $\$230{,}000$.
Part c)
Set $y = 200000$ and solve for $x$: \[200000 = 12x + 110000\] \[90000 = 12x\] \[x = \frac{90000}{12} = 7500\]
The company should spend $\$7{,}500$ on advertising.
Answer:
a) $y = 12x + 110{,}000$
b) Revenue is $\$230{,}000$
c) Spend $\$7{,}500$ on advertising
Verification: For $x = 5000$: $y = 12(5000) + 110000 = 60000 + 110000 = 170000$ ✓. For $x = 12000$: $y = 12(12000) + 110000 = 144000 + 110000 = 254000$ ✓
20) For problem 19, explain the following:
a. Explain what the slope of the line tells us about the effect on sales revenue of money spent on advertising. Be specific, explaining both the number and the sign of the slope in the context of this problem.
b. Explain what the y intercept of the line tells us about the sales revenue in the context of this problem.
Problem 20 Solution
Using the revenue function from Problem 19: $y = 12x + 110{,}000$.
Part a) Interpreting the slope:
The slope is $m = 12$, which is positive.
- The number: The value 12 means that for every additional $\$1$ spent on advertising, the company's sales revenue increases by $\$12$. Equivalently, every $\$1{,}000$ increase in advertising spending produces $\$12{,}000$ in additional revenue.
- The sign: The positive sign indicates a direct relationship — as advertising spending increases, sales revenue also increases. This makes intuitive sense: more advertising exposure leads to more sales.
- Practical meaning: The company gets a 12-to-1 return on its advertising investment. Each advertising dollar generates $\$12$ in revenue, suggesting that advertising is a worthwhile investment for this company.
Part b) Interpreting the y-intercept:
The y-intercept is $b = 110{,}000$.
- This represents the sales revenue when $x = 0$ — that is, when the company spends nothing on advertising.
- In context, $\$110{,}000$ is the company's base (organic) sales revenue that it would earn through existing brand recognition, word of mouth, repeat customers, and other non-advertising channels.
- This tells us that even without any advertising expenditure, ABC Co. can still expect $\$110{,}000$ in sales revenue.
Answer:
a) The slope of 12 (positive) means that for every additional dollar spent on advertising, sales revenue increases by $\$12$.
b) The y-intercept of $\$110{,}000$ represents the base sales revenue the company earns with zero advertising spending.
21) Mugs Café sells 1000 cups of coffee per week if it does not advertise. For every $\$50$ spent in advertising per week, it sells an additional 150 cups of coffee.
a) Find a linear function that gives $y =$ number of cups of coffee sold per week $x =$ amount spent on advertising per week.
b) How many cups of coffee does Mugs Café expect to sell if $\$100$ per week is spent on advertising?
Problem 21 Solution
Part a)
Step 1: Identify the y-intercept.
With no advertising ($x = 0$), the café sells 1000 cups per week, so $b = 1000$.
Step 2: Calculate the slope.
For every $\$50$ spent, 150 additional cups are sold:
\[m = \frac{150 \text{ cups}}{50 \text{ dollars}} = 3\]
This means each dollar of advertising generates 3 additional cups of coffee sold.
Step 3: Write the function: \[y = 3x + 1000\]
Part b)
Substitute $x = 100$: \[y = 3(100) + 1000 = 300 + 1000 = 1300\]
Answer:
a) $y = 3x + 1000$
b) Mugs Café expects to sell 1,300 cups of coffee per week.
Verification: For $x = 0$: $y = 3(0) + 1000 = 1000$ ✓ (matches "1000 cups with no advertising"). For $x = 50$: $y = 3(50) + 1000 = 1150 = 1000 + 150$ ✓ (150 additional cups for $\$50$ spent).
22) Party Sweets makes baked goods that can be ordered for special occasions. The price is $\$24$ to order one dozen (12 cupcakes) and $\$9$ for each additional 6 cupcakes.
a) Find a linear function that gives the total price of a cupcake order as a function of the number of cupcakes ordered.
b) Find the price for an order of 5 dozen (60) cupcakes.
Problem 22 Solution
Part a)
Step 1: Identify two data points. Let $x =$ number of cupcakes and $y =$ total price.
- 12 cupcakes (1 dozen) costs $\$24$: point $(12, 24)$
- 18 cupcakes (1 dozen + 6 additional) costs $\$24 + \$9 = \$33$: point $(18, 33)$
Step 2: Calculate the slope $m$: \[m = \frac{33 - 24}{18 - 12} = \frac{9}{6} = 1.50\]
Each cupcake costs $\$1.50$.
Step 3: Find the y-intercept $b$ using the point $(12, 24)$: \[24 = 1.50(12) + b\] \[24 = 18 + b\] \[b = 6\]
The $\$6$ represents a base order fee.
Step 4: Write the pricing function: \[y = 1.50x + 6\]
Part b)
Find the price for 60 cupcakes ($x = 60$): \[y = 1.50(60) + 6 = 90 + 6 = 96\]
Answer:
a) $y = 1.50x + 6$
b) An order of 5 dozen (60) cupcakes costs $\$96$.
Verification: For $x = 12$: $y = 1.50(12) + 6 = 18 + 6 = 24$ ✓. For $x = 18$: $y = 1.50(18) + 6 = 27 + 6 = 33$ ✓ (which is $\$24 + \$9$).