Problem 330 - Calculus Volume 1

Question

Consider a limousine that gets \(m(v)=\frac{(120-2 v)}{5} \mathrm{mi} / \mathrm{gal}\) at speed \(v, \quad\) the chauffeur costs \(\$ 15 / \mathrm{h},\) and gas is \(\$ 3.5 / \mathrm{gal}\). Find the cost per mile at speed \(v\).

Step-by-Step Solution

Answer

The cost per mile at speed \( v \) is \( C(v) = \frac{15}{v} + \frac{17.5}{120-2v} \text{ dollars}.\)

Step 1: Understand the Variables

The function given is the mileage function of the limousine, \( m(v) = \frac{(120 - 2v)}{5} \text{ mi/gal} \). It means the mileage depends on the speed \( v \) of the vehicle. The cost of the chauffeur is \( \\(15/\text{h} \) and the gas costs \( \\)3.5/\text{gal} \). Our task is to find the cost per mile of traveling at speed \( v \).

Step 2: Define the Cost Components

To find the cost per mile, we need to combine the costs of the chauffeur and the fuel used per mile. Let us first express the time taken to travel one mile at speed \( v \) (in hours), which is \( \frac{1}{v} \text{ hours/mile} \).

Step 3: Compute the Chauffeur Cost Per Mile

Given the chauffeur costs \( \$15/\text{h} \), the time to travel one mile at speed \( v \) is \( \frac{1}{v} \text{ hours} \). Thus, the cost per mile for the chauffeur is \( 15 \times \frac{1}{v} = \frac{15}{v} \text{ dollars/mile} \).

Step 4: Compute the Fuel Cost Per Mile

The limousine's mileage is given by \( m(v) = \frac{(120-2v)}{5} \). Therefore, the fuel used per mile is \( \frac{1}{m(v)} = \frac{5}{120-2v} \text{ gallons/mile} \). Given the fuel price is \( \$3.5/\text{gal} \), the fuel cost per mile becomes \( 3.5 \times \frac{5}{120-2v} = \frac{17.5}{120-2v} \text{ dollars/mile} \).

Step 5: Sum the Costs to Find Total Cost Per Mile

The total cost per mile is the sum of the chauffeur and fuel costs per mile. Therefore, the cost per mile \( C(v) \) is given by: \[C(v) = \frac{15}{v} + \frac{17.5}{120-2v}\]

Key Concepts

Mileage Function

The mileage function is a critical concept in understanding the relationship between speed and fuel efficiency. For the limousine, this function is given by:\[ m(v) = \frac{(120 - 2v)}{5} \text{ mi/gal} \]This formula indicates how many miles the limousine can travel per gallon of gas as a function of speed \( v \). As speed increases, the value of \( 2v \) also increases, which decreases the overall mileage. This decrease represents the general observation that vehicles often become less fuel-efficient at higher speeds due to increased air resistance and engine wear.

• At low speeds, the mileage is high because less energy is lost overcoming air resistance.
• As speed increases, the resistance grows, making the vehicle consume more fuel per mile traveled.

Understanding this trade-off helps in finding the optimal speed where fuel consumption is minimized without compromising too much on travel time.

Cost Analysis

Cost analysis in this context refers to calculating the expense incurred per mile when driving the limousine at a certain speed. The limousine's costs per mile are derived from two main components: the chauffeur's wage and the fuel cost.Firstly, the chauffeur's cost per mile is:\[ \frac{15}{v} \text{ dollars/mile} \]This result forms from knowing the chauffeur earns \\(15 per hour and considering the time to travel one mile is \( \frac{1}{v} \) hours. Thus, as speed increases, the time taken, and hence the chauffeur costs per mile, decreases.Secondly, the fuel cost per mile calculates as:\[ \frac{17.5}{120-2v} \text{ dollars/mile} \]This calculation arises from noting the fuel cost is \\)3.5 per gallon and the fuel efficiency is affected by mileage \( m(v) \). Combining these elements allows us to evaluate total costs effectively.

Speed Optimization

Speed optimization is essential to balance the costs and fuel efficiency effectively. By understanding the equation:\[ C(v) = \frac{15}{v} + \frac{17.5}{120-2v} \]we can find an optimal speed \( v \) where the cost per mile \( C(v) \) is minimized. Optimization in calculus involves finding where the derivative of the cost function equals zero, indicating a minimum or maximum point.

• Using calculus, differentiate \( C(v) \) with respect to speed \( v \).
• Set the derivative equal to zero to find critical points that might represent the minimum cost per mile.
• It's crucial to check these points to ensure they are minimum values and not maximum values.

Through speed optimization, we can suggest the most cost-effective speed for operating the limousine, taking into account both chauffeur and fuel costs. This approach builds a practical understanding of using calculus to solve real-world optimization problems.