Problem 72 - Algebra for College Students

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 \(C(m)=50+0.32 m\). Compute \(C(75)\), \(C(150), C(225)\), and \(C(650) . \quad \$ 74 ; \$ 98 ; \$ 122 ; \$ 258\)

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 is given by \( C(m) = 50 + 0.32m \), where \( m \) is the number of miles driven. This means that the total cost is the sum of a fixed daily fee and a variable fee that depends on the number of miles.

Step 2: Compute C(75)

Substitute \( m = 75 \) into the cost function: \[ C(75) = 50 + 0.32 \times 75 \] Calculate the product: \[ C(75) = 50 + 24 = 74 \] The cost for 75 miles is \$74.

Step 3: Compute C(150)

Substitute \( m = 150 \) into the cost function: \[ C(150) = 50 + 0.32 \times 150 \] Calculate the product: \[ C(150) = 50 + 48 = 98 \] The cost for 150 miles is \$98.

Step 4: Compute C(225)

Substitute \( m = 225 \) into the cost function: \[ C(225) = 50 + 0.32 \times 225 \] Calculate the product: \[ C(225) = 50 + 72 = 122 \] The cost for 225 miles is \$122.

Step 5: Compute C(650)

Substitute \( m = 650 \) into the cost function: \[ C(650) = 50 + 0.32 \times 650 \] Calculate the product: \[ C(650) = 50 + 208 = 258 \] The cost for 650 miles is \$258.

Key Concepts

Cost Function

A cost function is a mathematical formula that helps in determining the total cost associated with producing a certain number of goods or consuming a service. In our exercise, the car rental cost is modeled using a cost function \( C(m) = 50 + 0.32m \). Here, the function represents the total cost of renting a car based on the number of miles traveled, denoted by \( m \).

• The first part, \( 50 \), is a constant which indicates the fixed daily charge just for renting the car.
• The second part, \( 0.32m \), is the variable component that goes up with each mile driven.

This shows that in cost functions, you often have a fixed component and a variable component. Understanding this helps in anticipating the possible expenses involved. For practical applications, you just need to plug in the miles driven to find out your total cost. Breaking it down like this makes it easy to plan any budget-related decision such as trips.

Variables in Algebra

Variables in algebra are symbols that represent numbers or values that can change. In our exercise, the variable \( m \) represents the number of miles driven during the car rental period. Using variables offers flexibility and enables you to create general formulas that work for various inputs.Let's explore this further:

• Variables allow us to solve problems across different scenarios without rewriting equations repeatedly.
• They serve as placeholders, enabling substitution and thus calculation for specific instances, as seen when calculating \( C(75), C(150), C(225), \) and \( C(650) \).

Variables make mathematical problems scalable and adaptable. So, grasping this concept is key in unlocking a wide array of algebra problems. With variables, you broaden your mathematical toolkit.

Mathematical Modeling

Mathematical modeling is the process of representing real-world scenarios with mathematical expressions and equations. This is a powerful tool, as it allows us to use mathematics to simulate and predict behaviors and outcomes. In our situation, the cost of renting a car over a certain distance is modeled by the function \( C(m) = 50 + 0.32m \).Why is this useful?

• It provides a clear, calculable method to determine costs based on easily measurable data (distance driven).
• By turning a real-world problem into a mathematical model, you can apply logic and problem-solving skills to derive solutions efficiently.

Modeling with mathematics simplifies complex real-life problems by breaking them down into manageable parts. This is crucial in fields ranging from engineering and physics to economics and finance. It's all about taking something complicated and turning it into something we can understand and analyze easily. Being able to see things mathematically gives you a new perspective on a wide range of problems.