Problem 26 - Algebra for College Students

Question

Suppose that a car rental agency charges a fixed amount per day plus an amount per mile for renting a car. Heidi rented a car one day and paid \(\$ 80\) for 200 miles. On another day she rented a car from the same agency and paid \(\$ 117.50\) for 350 miles. Determine the linear function that the agency could use to determine its daily rental charges. \(f(x)=0.25 x+30\)

Step-by-Step Solution

Answer

The linear function is \( f(x) = 0.25x + 30 \).

Step 1: Define Variables and Set Up Equations

Let's denote the fixed daily rate as \( d \) and the charge per mile as \( m \). Using the information given, we can create two equations. For the first day Heidi paid \( \\(80 \) for 200 miles, which gives us the equation: \( d + 200m = 80 \). For the second day, she paid \( \\)117.50 \) for 350 miles: \( d + 350m = 117.50 \).

Step 2: Solve the System of Equations

The system of equations is:\( \begin{cases} d + 200m = 80 \cr d + 350m = 117.50 \end{cases} \). To eliminate \( d \), subtract the first equation from the second: \( (d + 350m) - (d + 200m) = 117.50 - 80 \). Simplifying gives \( 150m = 37.50 \).

Step 3: Calculate the Charge per Mile

To find \( m \), divide both sides of the equation obtained in Step 2 by 150: \( m = \frac{37.50}{150} = 0.25 \). This means the agency charges \( \$0.25 \) per mile.

Step 4: Find the Fixed Daily Rate

Substitute \( m = 0.25 \) back into one of the original equations, for example, \( d + 200(0.25) = 80 \). Simplify to find \( d + 50 = 80 \). Subtract 50 from both sides to get \( d = 30 \). This means the fixed daily charge is \( \$30 \).

Step 5: Write the Linear Function

Now combine \( d = 30 \) and \( m = 0.25 \) to write the linear function for the cost: \( f(x) = 0.25x + 30 \). This function represents the total cost \( f(x) \) of renting a car for \( x \) miles in a day.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. In this particular problem, we have two equations stemming from Heidi's two separate car rentals.
These equations are built upon the given costs and distances:

• First equation: She paid \\(80 for 200 miles, represented as \( d + 200m = 80 \).
• Second equation: She paid \\)117.50 for 350 miles, shown as \( d + 350m = 117.50 \).

The goal here is to find the values of \( d \) (daily fixed charge) and \( m \) (mileage charge). By solving the system, we discover the cost structure. To do this, we align and subtract the equations to eliminate one variable, allowing us to solve for the other.
This technique is fundamental when dealing with problems involving multiple scenarios where a consistent rule applies.

Fixed and Variable Costs

In many business models, costs are divided into fixed and variable components. Fixed costs are constant, no matter how much you use or produce, while variable costs fluctuate depending on usage or production.
In our car rental scenario:

• Fixed costs are represented by the daily charge \( d \), which remains constant regardless of the number of miles driven.
• Variable costs are represented by \( m \), the cost per mile driven.

Understanding these components allows businesses and consumers to predict expenses more accurately. Fixed costs provide stability, while variable costs account for additional usage.

Cost Function

A cost function in mathematical terms is a linear expression that describes how total cost changes with the level of activity or production. For Heidi’s car rental, the cost function is expressed as:
\( f(x) = 0.25x + 30 \)
Here, \( f(x) \) represents the total daily charge based on mileage \( x \).

• The term \( 0.25x \) represents the variable cost depending on the miles driven.
• The constant \( 30 \) is the fixed daily rate.

This function enables potential customers to calculate their expected expenses based on anticipated mileage, making financial planning more transparent.

Algebraic Problem Solving

Algebraic problem solving involves using mathematical techniques to find unknown values. In this context, the unknown values are the fixed rate \( d \) and the mileage rate \( m \).
The step-by-step approach involves:

• Setting up equations based on the problem statement.
• Using algebraic manipulation to isolate and solve for each variable.
• Substituting back to verify the solution.

In real-world scenarios, algebra helps break down complex situations into manageable parts, allowing for systematic investigation and solution finding.