Problem 69 - Intermediate Algebra

Question

A car rental agency charges \(\$ 50\) per day plus \(\$ 0.32\) a mile. Therefore, the daily charge for renting a car is a function of the number of miles traveled \((m)\) and can be expressed as \(C(\mathrm{~m})=50+0.32 \mathrm{~m}\). Compute \(C(75)\), \(C(150), C(225)\), and \(C(650)\).

Step-by-Step Solution

Answer

\( C(75) = 74, C(150) = 98, C(225) = 122, C(650) = 258. \)

Step 1: Understand the Function

The cost function provided is \( C(m) = 50 + 0.32m \), where \( C(m) \) represents the total rental cost based on the number of miles \( m \). We need to compute \( C(m) \) for different values of \( m \): 75, 150, 225, and 650.

Step 2: Substitute and Calculate for m = 75

Substitute \( m = 75 \) into the function: \( C(75) = 50 + 0.32 \times 75 \). Calculate the multiplication: \( 0.32 \times 75 = 24 \). Thus, \( C(75) = 50 + 24 = 74 \).

Step 3: Substitute and Calculate for m = 150

Substitute \( m = 150 \) into the function: \( C(150) = 50 + 0.32 \times 150 \). Calculate the multiplication: \( 0.32 \times 150 = 48 \). Thus, \( C(150) = 50 + 48 = 98 \).

Step 4: Substitute and Calculate for m = 225

Substitute \( m = 225 \) into the function: \( C(225) = 50 + 0.32 \times 225 \). Calculate the multiplication: \( 0.32 \times 225 = 72 \). Thus, \( C(225) = 50 + 72 = 122 \).

Step 5: Substitute and Calculate for m = 650

Substitute \( m = 650 \) into the function: \( C(650) = 50 + 0.32 \times 650 \). Calculate the multiplication: \( 0.32 \times 650 = 208 \). Thus, \( C(650) = 50 + 208 = 258 \).

Key Concepts

Function Evaluation

Function evaluation involves determining the output of a function for specific inputs. Here, the cost function is expressed as:

• \(C(m) = 50 + 0.32m\)

This function estimates the daily cost of renting a car based on the miles traveled, \(m\). The term \(50\) stands for the basic daily charge, while \(0.32m\) is the added charge per mile.
Evaluation involves inserting specific values into the function. For example, to find \(C(75)\), substitute \(75\) into \(m\) in the function equation. Then, compute the result. This is repeated for other values like \(150\), \(225\), and \(650\). This step allows you to calculate the total cost for each specific number of miles traveled.

Algebraic Operations

Algebraic operations enable us to manipulate and simplify equations. Understanding these operations is crucial when working with cost functions or similar expressions.
To find the cost function's output for different values of miles \(m\), use basic multiplication and addition:

• Multiply \(0.32\) by the number of miles \(m\) traveled.
• Add the result of \(0.32m\) to the fixed daily charge of \(50\).

For instance, to compute \(C(75)\), start with multiplying:

• \(0.32 \times 75 = 24\)

Then add this to the fixed cost:

• \(50 + 24 = 74\)

Repeatedly performing these operations with each given \(m\) value lets you determine the total cost effortlessly.

Word Problems

Word problems require translating textual information into mathematical expressions and solving the resulting equations. In this rental car scenario, the problem describes a situation where you compute the cost based on daily and mileage charges.
To solve such problems, first, identify crucial elements:

• The fixed cost per day (\\(50)
• The variable cost per mile (\\)0.32 per mile)

Next, develop an equation to model these findings, like \(C(m) = 50 + 0.32m\).
The last step involves evaluating this function with specific values provided in the problem statement. This method helps transform a real-world situation into a mathematical problem that can be solved through known algebraic techniques.