Problem 1 - Applied Calculus

Question

Concern the cost, \(C\), of renting a car from a company which charges $$\$ 40$$ a day and 15 cents a mile, so \(C=f(d, m)=40 d+0.15 m\), where \(d\) is the number of days, and \(m\) is the number of miles. Make a table of values for \(C\), using \(d=1,2,3,4\) and \(m=100,200,300,400\). You should have 16 values in your table.

Step-by-Step Solution

Answer

Fill the table with costs: \( d = 1 \) to \( 4 \), and \( m = 100 \) to \( 400 \). Costs range from 55 to 220.

Step 1: Understand the Formula

The cost \( C \) of renting the car is calculated using the formula \( C = 40d + 0.15m \), where \( d \) represents the number of days the car is rented, and \( m \) represents the number of miles driven. We'll calculate \( C \) for each combination of the given \( d \) and \( m \) values.

Step 2: Create a Table Framework

We'll create a table with rows representing the number of days \( d = 1, 2, 3, 4 \) and columns representing the number of miles \( m = 100, 200, 300, 400 \). We will fill in each cell with the computed cost \( C \) using the formula provided.

Step 3: Calculate Costs for \(d = 1\)

For \( d = 1 \), calculate \( C \) for each \( m \) value:1. \( m = 100: C = 40(1) + 0.15(100) = 40 + 15 = 55 \)2. \( m = 200: C = 40(1) + 0.15(200) = 40 + 30 = 70 \)3. \( m = 300: C = 40(1) + 0.15(300) = 40 + 45 = 85 \)4. \( m = 400: C = 40(1) + 0.15(400) = 40 + 60 = 100 \)

Step 4: Calculate Costs for \(d = 2\)

For \( d = 2 \), calculate \( C \) for each \( m \) value:1. \( m = 100: C = 40(2) + 0.15(100) = 80 + 15 = 95 \)2. \( m = 200: C = 40(2) + 0.15(200) = 80 + 30 = 110 \)3. \( m = 300: C = 40(2) + 0.15(300) = 80 + 45 = 125 \)4. \( m = 400: C = 40(2) + 0.15(400) = 80 + 60 = 140 \)

Step 5: Calculate Costs for \(d = 3\)

For \( d = 3 \), calculate \( C \) for each \( m \) value:1. \( m = 100: C = 40(3) + 0.15(100) = 120 + 15 = 135 \)2. \( m = 200: C = 40(3) + 0.15(200) = 120 + 30 = 150 \)3. \( m = 300: C = 40(3) + 0.15(300) = 120 + 45 = 165 \)4. \( m = 400: C = 40(3) + 0.15(400) = 120 + 60 = 180 \)

Step 6: Calculate Costs for \(d = 4\)

For \( d = 4 \), calculate \( C \) for each \( m \) value:1. \( m = 100: C = 40(4) + 0.15(100) = 160 + 15 = 175 \)2. \( m = 200: C = 40(4) + 0.15(200) = 160 + 30 = 190 \)3. \( m = 300: C = 40(4) + 0.15(300) = 160 + 45 = 205 \)4. \( m = 400: C = 40(4) + 0.15(400) = 160 + 60 = 220 \)

Step 7: Fill the Table with Values

Using the calculations from Steps 3 to 6, fill the table with the computed costs \( C \) as follows: \[\begin{array}{c|c|c|c|c}& m=100 & m=200 & m=300 & m=400 \\hline d=1 & 55 & 70 & 85 & 100 \ d=2 & 95 & 110 & 125 & 140 \ d=3 & 135 & 150 & 165 & 180 \ d=4 & 175 & 190 & 205 & 220 \\end{array}\]

Key Concepts

Cost Calculation

Understanding cost calculation is about determining the total expense based on various influencing factors. In this exercise, we're looking at the cost associated with renting a car. The formula given is \( C = 40d + 0.15m \), where \( C \) represents the total cost, \( d \) is the number of days, and \( m \) is the number of miles driven.
Here's how the formula breaks down:

• For each day the car is rented, \( \\(40 \) is added to the cost, so \( 40d \) gives the daily rental cost.
• Every mile driven adds \( \\)0.15 \) to the cost, hence \( 0.15m \) covers the mileage cost.

Plugging values into the formula allows us to calculate the exact cost for different combinations of days and miles. This approach highlights the practical application of linear functions in real-world scenarios, helping us make informed financial decisions.

Function Tables

Function tables are fantastic tools that help organize data clearly and understandably. In this exercise, the table is designed to show how costs change with different inputs of days and miles.
The table setup involves selecting a range of values for \( d \) (days) and \( m \) (miles) to systematically explore various cost scenarios:

• Days (\( d \)): 1, 2, 3, and 4
• Miles (\( m \)): 100, 200, 300, and 400

By computing the cost \( C \) for each pair of \( d \) and \( m \), we generate distinct values filling the table. This organized presentation facilitates spotting trends, making it easier for students to understand how each variable affects the total cost. Function tables simplify the manipulation and visualization of data, aiding in grasping how input changes influence outcomes.

Applied Mathematics

Applied mathematics involves using mathematical methods to solve practical problems. In this context, we are applying a linear function to predict costs associated with car rentals. This application demystifies how mathematical formulas operate in everyday life.
Knowing how to apply these calculations can help in various professional and personal scenarios. Here's why it matters:

• It empowers individuals to forecast expenses accurately, leading to better budgeting.
• It offers insights into pricing strategies used by businesses, opening doors to negotiate better deals.
• Such exercises harness algebraic skills to solve practical challenges, reinforcing concepts learned in classrooms.

Applying the learned methods to similar problems enhances understanding and builds confidence. As students become more comfortable using these mathematical tools, they can more easily transition from theoretical learning to its practical application in the real world.