Problem 13 - Applied 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

Company 1 is cheaper for <200 miles; company 2 for >200 miles.

Step 1: Understanding the Cost Formulas

To find the formula for the cost as a function of distance traveled, we will define variables: let \( x \) be the number of miles driven. For the first company, the cost per day is \( \\(40 \) plus \( 0.15x \). For the second company, the cost is \( \\)50 \) plus \( 0.10x \).

Step 2: Formulating Cost Equations

For the first company, the cost function is \( C_1(x) = 40 + 0.15x \). For the second company, the cost function is \( C_2(x) = 50 + 0.10x \).

Step 3: Constructing the Cost Graph

Plot the cost functions \( C_1(x) = 40 + 0.15x \) and \( C_2(x) = 50 + 0.10x \) on a graph with \( x \)-axis representing miles driven and \( y \)-axis representing total cost. Each line's y-intercept is at 40 and 50 respectively, with respective slopes of 0.15 and 0.10.

Step 4: Determining Cheaper Company by Intersection Point

Find the intersection point of the two lines to determine the mileage at which the cost is the same: \( 40 + 0.15x = 50 + 0.10x \). Subtract 40 from both sides: \( 0.15x = 10 + 0.10x \). Simplify by subtracting \( 0.10x \) from both sides: \( 0.05x = 10 \). Solving for \( x \), we find \( x = 200 \) miles.

Step 5: Concluding Decision Criteria

The first company is cheaper if fewer than 200 miles are driven, as \( C_1(x) < C_2(x) \). The second company is preferable if more than 200 miles are driven, as \( C_2(x) < C_1(x) \).

Key Concepts

Cost Function

In applied calculus, a cost function models the total cost associated with an activity, in this case, renting cars. The cost depends on two factors here: a fixed cost per day and a variable cost based on miles driven. To represent this mathematically, we use a linear cost function.

• For the first company, every day costs \(40 regardless of mileage, and it charges 15 cents per mile. The formula is: \( C_1(x) = 40 + 0.15x \) where \( x \) is the number of miles driven.
• The second company has a slightly different structure. It charges \)50 per day and 10 cents per mile. Thus, its cost function is: \( C_2(x) = 50 + 0.10x \).

These formulas help predict costs based on distance traveled. They break the cost into initial fixed costs and ongoing variable costs. Understanding each component aids in comparing different options effectively.

Graphing Linear Equations

Graphing linear equations is a powerful tool for visualizing cost functions. It enables you to see how costs increase relative to miles driven. Each cost function can be represented as a straight line on a graph.

• The y-intercept represents the fixed daily cost. For example, the line for the first company starts at 40 on the y-axis, while the second company's line starts at 50.
• The slope indicates how fast costs grow per mile. A slope of 0.15 and 0.10 reflects this growth rate for the first and second companies respectively.

By plotting these lines on a graph, you can easily compare companies. The intersection point of these lines indicates where the cost of both options is equal. Beyond this point, one line will be consistently higher or lower, predicting the cheaper option for future miles.

Optimization Problem

In this context, the optimization problem involves determining which car rental option costs less based on miles driven. This requires solving for the point at which both cost functions are equal, known as the intersection point.

• Set the two cost functions equal: \( 40 + 0.15x = 50 + 0.10x \).
• Simplify and solve for \( x \), yielding \( x = 200 \) miles.
• This tells us that up to 200 miles, the first company is cheaper. Beyond 200 miles, the second company becomes the economical choice.

This process highlights the power of applied calculus in real-life decision making. Understanding where cost functions intersect helps in choosing the most financially favorable option, effectively addressing the optimization problem at hand.