Problem 43 - Calculus

Question

A company rents cars at \(\$ 40\) a day and 15 cents a mile. Its competitor's cars are \(\$ 50\) a day and 10 cents a mile. (a) For each company, give a formula for the cost of renting a car for a day as a function of the distance traveled. (b) On the same axes, graph both functions. (c) How should you decide which company is cheaper?

Step-by-Step Solution

Answer

Use Company 1 for trips under 200 miles, and Company 2 for trips over 200 miles.

Step 1: Define Variables

Let \(d\) represent the number of miles driven in a day. We will use \(C_1\) for the cost of the first company and \(C_2\) for the cost of the competitor. We will express both costs as functions of \(d\).

Step 2: Formulate Costs for Company 1

For the first company, the daily rental cost is \$40 and the cost per mile is 15 cents. Thus, the cost function is:\[ C_1(d) = 40 + 0.15d \]

Step 3: Formulate Costs for Company 2

For the competitor, the daily rental cost is \$50 and the cost per mile is 10 cents. Thus, the cost function is:\[ C_2(d) = 50 + 0.10d \]

Step 4: Plot Both Cost Functions

To graph these functions, plot \(C_1(d) = 40 + 0.15d\) and \(C_2(d) = 50 + 0.10d\) on the same axes with \(d\) on the x-axis and cost \(C\) on the y-axis.

Step 5: Analyze the Graph

Find the intersection point by setting \(C_1(d) = C_2(d)\):\[ 40 + 0.15d = 50 + 0.10d \]Solve for \(d\):\[ 0.05d = 10 \] \[ d = 200 \]The intersection point is at \(d = 200\). Below 200 miles, Company 1 is cheaper, while above 200 miles, Company 2 is cheaper.

Key Concepts

Linear Equations

Linear equations are essential building blocks in mathematics, especially when analyzing cost functions in real-life scenarios. In this exercise, linear equations represent the cost of renting a car based on the number of miles driven. The formula for the cost combines a fixed daily rate and a variable component based on mileage.

For example, the cost function for the first company is given by the equation \(C_1(d) = 40 + 0.15d\). Here:

• \(40\) is the fixed rate in dollars - the cost without driving any miles.
• \(0.15d\) is the variable cost, where \(d\) denotes miles driven and \(0.15\) stands for the cost per mile.

Similarly, the competitor company's cost function is \(C_2(d) = 50 + 0.10d\). Understanding how these components interact within a linear equation enables us to model and analyze different cost scenarios accurately.

Graphing Functions

Graphing functions allows for a visual representation of how costs change with distance. This exercise involves graphing the cost functions \(C_1(d) = 40 + 0.15d\) and \(C_2(d) = 50 + 0.10d\) on a coordinate plane.

When graphing, use the following steps:

• Select values for \(d\), the number of miles driven. Typically start with low mileages and increase incrementally.
• Calculate the corresponding values of \(C_1(d)\) and \(C_2(d)\) using these \(d\) values.
• Plot these points on a graph with \(d\) on the x-axis and cost \(C\) on the y-axis.
• Connect the points with straight lines, as the functions are linear.

This graph helps identify the rate at which costs differ between companies as mileage increases. The intersection point of the two graphs indicates where the costs equalize.

Comparative Analysis

A comparative analysis allows us to decide which rental company is more economical. By analyzing the given functions and their graphs, we can determine which option is cheaper based on mileage.

Consider the intersection of the graphs of \(C_1(d) = 40 + 0.15d\) and \(C_2(d) = 50 + 0.10d\) by setting them equal: \\[ 40 + 0.15d = 50 + 0.10d \]Solving this equation, we find that both companies break even at 200 miles, with the solution \(d = 200\):

• Below 200 miles, Company 1 is cheaper because its graph lies below that of Company 2.
• Above 200 miles, Company 2 is more cost-effective since its graph is lower.

By understanding these comparisons, we can make informed decisions about which rental car company to choose based on our travel plans.