Problem 18 - Applied Calculus

Question

The cost of renting a car from a certain company is $$\$ 40$$ per day plus 15 cents per mile, and so we have $$C=40 d+0.15 m$$ Find \(\partial C / \partial d\) and \(\partial C / \partial m\). Give units and explain why your answers make sense.

Step-by-Step Solution

Answer

\(\partial C/\partial d = 40\) dollars per day; \(\partial C/\partial m = 0.15\) dollars per mile.

Step 1: Identify Variables in the Cost Function

The cost function is given by the equation \(C = 40d + 0.15m\), where \(d\) represents the number of days the car is rented, and \(m\) represents the number of miles driven. We need to find the partial derivatives \(\partial C / \partial d\) and \(\partial C / \partial m\) to understand how the cost changes with respect to days and miles, respectively.

Step 2: Calculate \(\partial C / \partial d\)

The partial derivative \(\partial C / \partial d\) represents the rate of change of the cost \(C\) with respect to the number of rental days \(d\), keeping the miles \(m\) constant. By differentiating \(C = 40d + 0.15m\) with respect to \(d\), we find \(\partial C / \partial d = 40\). This means the cost increases by \$40 for each additional day the car is rented.

Step 3: Calculate \(\partial C / \partial m\)

The partial derivative \(\partial C / \partial m\) represents the rate of change of the cost \(C\) with respect to the number of miles \(m\), holding the rental days \(d\) constant. By differentiating \(C = 40d + 0.15m\) with respect to \(m\), we find \(\partial C / \partial m = 0.15\). This signifies that the cost increases by $0.15 or 15 cents for each additional mile driven.

Key Concepts

Cost Function

The cost function is a mathematical expression used to represent how costs vary with factors or inputs. In this exercise, the cost function is represented by the equation \(C = 40d + 0.15m\). Here, \(C\) is the total cost of renting a car, \(d\) stands for the number of days the car is rented, and \(m\) symbolizes the number of miles driven.

Understanding a cost function is crucial in real-life settings where we want to predict the cost based on certain changes, like how many days you need a rental car or how far you plan to drive. It simplifies complex situations by giving us a clear formula to calculate costs easily.

The components of this function include:

• \(40d\) which signifies the cost component per day. For each day, an additional \$40 is charged.
• \(0.15m\) which represents the mileage charge, where 15 cents is added for each mile driven.

By having this function, you can plan and budget effectively by simply plugging in your days and miles values to forecast the total cost.

Rate of Change

The rate of change is a concept in mathematics that helps us understand how a quantity changes in response to another variable. When talking about the cost function \(C = 40d + 0.15m\), we're interested in how the cost \(C\) changes as we adjust the number of days \(d\) or miles \(m\).

In this scenario, we use partial derivatives to find the rate of change.

• \(\partial C / \partial d\): This partial derivative measures how the cost changes with each additional rental day while keeping the number of miles constant. Here, it's \(40\), indicating the cost increases by \\(40 per extra day.
• \(\partial C / \partial m\): This represents the change in cost for each extra mile driven, with the days constant. It's \(0.15\) or 15 cents per mile.

Understanding these rates helps make informed decisions. For instance, if you know every additional day adds \\)40 to your bill, it makes it clear how much extending the car rental duration affects the overall cost.

Calculus Concepts

Calculus introduces us to powerful tools like derivatives, which allow us to analyze and understand continual change. In this context, we specifically use partial derivatives, focusing on their application in analyzing cost changes.

Partial derivatives let us study how a function changes if we alter just one variable while holding others constant. It essentially breaks down complex functions by examining each influencing factor individually. This technique is especially useful in multidimensional scenarios like our car rental problem, where both days and miles affect the cost.

Here's why partial derivatives are essential:

• They provide specific insights by isolating effects, aiding precise decision-making.
• They are instrumental in optimizing functions and solving real-world problems by pointing out the impact of each variable on the outcome.

By understanding these concepts, calculus becomes a valuable tool for real-world analysis, allowing us to make more calculated and well-informed choices.