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

The table contains 16 calculated costs, based on days and miles used.

Step 1: Understand the Formula

The cost to rent a car is described by the formula \(C = 40d + 0.15m\). This includes a daily charge of \(40 per day (`d`) and an additional cost of \)0.15 per mile (`m`).

Step 2: Set Up the Table Axes

We need to create a table with rows representing the number of days \(d = 1, 2, 3, 4\) and columns representing the number of miles \(m = 100, 200, 300, 400\).

Step 3: Calculate Cost for Each Combination

For each combination of \(d\) and \(m\), compute the cost using the formula \(C = 40d + 0.15m\).

Step 4: Compute and Fill the Table

Compute the cost for each \((d, m)\) combination:- For \(d=1, m=100\) to \(d=1, m=400\) * \(C = 40*1 + 0.15*100 = 55\) * \(C = 40*1 + 0.15*200 = 70\) * \(C = 40*1 + 0.15*300 = 85\) * \(C = 40*1 + 0.15*400 = 100\)- For \(d=2, m=100\) to \(d=2, m=400\) * \(C = 40*2 + 0.15*100 = 95\) * \(C = 40*2 + 0.15*200 = 110\) * \(C = 40*2 + 0.15*300 = 125\) * \(C = 40*2 + 0.15*400 = 140\)- For \(d=3, m=100\) to \(d=3, m=400\) * \(C = 40*3 + 0.15*100 = 135\) * \(C = 40*3 + 0.15*200 = 150\) * \(C = 40*3 + 0.15*300 = 165\) * \(C = 40*3 + 0.15*400 = 180\)- For \(d=4, m=100\) to \(d=4, m=400\) * \(C = 40*4 + 0.15*100 = 175\) * \(C = 40*4 + 0.15*200 = 190\) * \(C = 40*4 + 0.15*300 = 205\) * \(C = 40*4 + 0.15*400 = 220\)Fill these values systematically in the tabular format.

Key Concepts

Cost Calculation

Cost calculation is essential in determining the total expenses involved in a particular activity. For renting a car, as in our exercise, the cost depends on two main factors: the number of days you rent the car and the number of miles you drive.

The formula given is:

• \(C = 40d + 0.15m\)

Here, \(C\) represents the total cost, \(d\) symbolizes the number of days, and \(m\) stands for the miles driven.

• The first part, \(40d\), is the cost attributed to the days. It's a fixed cost, meaning each day adds \(\\(40\) regardless of miles driven.
• The second part, \(0.15m\), is the variable cost based on mileage driven. For every mile, you add \(\\)0.15\) to your cost.

Together, these two components form a linear equation that makes calculating rental costs straightforward. Understanding each part of this equation helps in predicting expenses accurately.

Table of Values

Creating a table of values provides a visual representation that helps us understand how costs change with different variables. It's like mapping out all possible outcomes based on various inputs.

In our case, the rows of the table represent different numbers of days \((d = 1, 2, 3, 4)\), while the columns represent different numbers of miles \((m = 100, 200, 300, 400)\). For each combination of \(d\) and \(m\), we calculate \(C\) using the formula mentioned earlier.

This table aids in quickly assessing how the total cost increases or decreases when either the days or miles change. It’s an excellent way to organize data for easy analysis and comparison. You can spot trends and make informed decisions, such as determining the optimal number of days to rent the car or how many miles to drive to stay within budget.

Function Evaluation

Function evaluation involves substituting specific values into a mathematical function, which in our scenario, is the cost function \(C = 40d + 0.15m\). With function evaluation, we determine the exact cost for given values of \(d\) and \(m\).

To do this:

• Substitute the given \(d\) value into the function.
• Substitute the given \(m\) value into the function.
• Perform the arithmetic operations to solve for \(C\).

For instance, if you have \(d=2\) and \(m=300\), you plug these numbers into the formula:

• \(C = 40 \times 2 + 0.15 \times 300\)
• This simplifies to \(C = 80 + 45\)
• Therefore, \(C = 125\)

Function evaluation enables precise calculation of cost based on selected range values, effectively transforming abstract mathematical models into practical solutions. This approach provides clarity on the cost impacts of varying rental conditions.