Problem 2
Question
Concern the cost, \(C,\) of renting a car from a company which charges \(\$ 40\) a day and 15 cents a mile, so \(C=f(d, m)=40 d+0.15 m,\) where \(d\) is the number of days, and \(m\) is the number of miles. (a) Find \(f(3,200)\) and interpret it. (b) Explain the significance of \(f(3, m)\) in terms of rental car costs. Graph this function, with \(C\) as a function of \(m\). (c) Explain the significance of \(f(d, 100)\) in terms of rental car costs. Graph this function, with \(C\) as a function of \(d\).
Step-by-Step Solution
Verified Answer
(a) \$150; (b) Cost varies by miles, graph: linear \( C = 120 + 0.15m \); (c) Cost varies by days, graph: linear \( C = 40d + 15 \).
1Step 1: Calculate f(3,200)
To find the cost when renting the car for 3 days and driving 200 miles, substitute \( d = 3 \) and \( m = 200 \) into the equation: \[ C = 40d + 0.15m \] Substitute to get: \[ C = 40(3) + 0.15(200) = 120 + 30 = 150 \] Thus, \( f(3,200) = 150 \). This means it costs \$150 to rent the car for 3 days and drive 200 miles.
2Step 2: Interpret f(3,m)
The function \( f(3, m) = 40(3) + 0.15m \) simplifies to \( C = 120 + 0.15m \). This expression shows the cost \( C \) as a function of miles \( m \), when renting the car for 3 days. The term \( 120 \) is the fixed cost for 3 days, and the term \( 0.15m \) is the variable cost, depending on the miles driven.
3Step 3: Graph f(3,m)
This graph will plot \( C = 120 + 0.15m \), with \( m \) on the x-axis and \( C \) on the y-axis. The y-intercept (at \( m = 0 \)) is \( 120 \), reflecting the cost for 3 days without any miles driven. The slope \( 0.15 \) represents the increase in cost per mile driven.
4Step 4: Interpret f(d,100)
The function \( f(d, 100) = 40d + 0.15(100) \) simplifies to \( C = 40d + 15 \). This expression reflects the cost \( C \) as a function of days \( d \), with a constant cost of \$15 for driving 100 miles. The term \( 40d \) is the daily rental cost, and \( 15 \) is a fixed mileage cost.
5Step 5: Graph f(d,100)
Graph \( C = 40d + 15 \), with \( d \) on the x-axis and \( C \) on the y-axis. The y-intercept (at \( d = 0 \)) is \( 15 \), representing the cost for driving 100 miles without any days. The slope \( 40 \) indicates the increase in cost per day rented.
Key Concepts
Cost FunctionGraphing FunctionsVariable Costs
Cost Function
When renting anything, understanding the **cost function** is crucial. In the car rental example, the cost function is used to calculate the total rental cost based on the number of days the car is used and the total miles driven. The formula given is:
In this case, the fixed cost is \( 40d \) for each day rented. There is also a variable cost, \( 0.15m \), which depends on the number of miles driven.
This function is beneficial because it allows you to predict and calculate total costs based on desired rental days and mileage.
- \( C = 40d + 0.15m \)
- Where \( C \) is total cost, \( d \) is number of days, \( m \) is number of miles.
In this case, the fixed cost is \( 40d \) for each day rented. There is also a variable cost, \( 0.15m \), which depends on the number of miles driven.
This function is beneficial because it allows you to predict and calculate total costs based on desired rental days and mileage.
Graphing Functions
Graphing functions helps visualize how different variables affect outcomes. In our example, you graph the cost function where:
The slope, 0.15, shows how cost increases relative to additional miles.
Similarly, plotting \( C = 40d + 15 \) with days, \( d \), on the x-axis introduces us to another perspective:
mis graphically depicted on the x-axis.C, representing the total cost, is shown on the y-axis.
The slope, 0.15, shows how cost increases relative to additional miles.
Similarly, plotting \( C = 40d + 15 \) with days, \( d \), on the x-axis introduces us to another perspective:
- The y-intercept may start at 15, meaning even without renting days, this base cost exists for any driving done (in our exercise, 100 miles).
- The slope is steeper at \( 40 \), showing costs rise faster as more days the car is rented.
Variable Costs
Variable costs change with the level of production or usage. In the discussed scenario, the variable cost is directly related to the number of miles driven.
If you plan to travel long distances, it's crucial to factor in these costs, as they can accumulate quickly.
For optimal budgeting, contrasting fixed and variable costs through these calculations helps in making informed renting decisions.
- It's represented as \( 0.15m \) in the cost function.
- For every mile driven, the cost increases by 15 cents.
If you plan to travel long distances, it's crucial to factor in these costs, as they can accumulate quickly.
For optimal budgeting, contrasting fixed and variable costs through these calculations helps in making informed renting decisions.
Other exercises in this chapter
Problem 2
Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$f(x, y)=x^{2}+4 x y, \quad x+y=100$$
View solution Problem 2
Find the partial derivatives in Problems. The variables are restricted to a domain on which the function is defined. $$f_{x} \text { and } f_{y} \text { if } f(
View solution Problem 3
Use Lagrange multipliers to find the maximum or minimum values of \(f(x, y)\) subject to the constraint. $$f(x, y)=x y, \quad 5 x+2 y=100$$
View solution Problem 3
Find the partial derivatives in Problems. The variables are restricted to a domain on which the function is defined.\(f_{x}\) and \(f_{y}\) if $f(x, y)=100 x^{2
View solution