Problem 75 - Algebra and Trigonometry

Question

cost of Driving The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May her driving cost was \(\$ 380\) for 480 \(\mathrm{mi}\) and in June her cost was \(\$ 460\) for 800 \(\mathrm{mi}\) . Assume that there is a linear relationship between the monthly cost \(C\) of driving a car and the distance driven \(d\) (a) Find a linear equation that relates \(C\) and \(d\) (b) Use part (a) to predict the cost of driving 1500 \(\mathrm{mi}\) per month. (c) Draw the graph of the linear equation. What does the slope of the line represent? (d) What does the \(y\) -intercept of the graph represent? (d) Why is a linear relationship a suitable model for this situation?

Step-by-Step Solution

Answer

(a) \( C = \frac{1}{4}d + 260 \), (b) \( \$635 \), (c) Slope: cost per mile, (d) Y-intercept: fixed cost, (e) Costs rise uniformly with distance.

Step 1: Determine the Variables

The monthly cost of driving is denoted by \( C \), and the distance driven is denoted by \( d \). We have two data points: May (\(C = 380, d = 480\)) and June (\(C = 460, d = 800\)).

Step 2: Find the Slope of the Line

To find the slope \( m \) of the line that relates \( C \) and \( d \), use the formula: \[m = \frac{C_2 - C_1}{d_2 - d_1} = \frac{460 - 380}{800 - 480} = \frac{80}{320} = \frac{1}{4}.\] This means the slope \( m \) is \( \frac{1}{4} \).

Step 3: Write the Linear Equation

Use the linear equation form \( C = md + b \). With \( m = \frac{1}{4} \), substitute one of the points, for example (480, 380), into the equation to find \( b \):\[ 380 = \frac{1}{4}(480) + b ote\rightarrow b = 380 - 120 = 260. \]Thus, the linear equation is \( C = \frac{1}{4}d + 260 \).

Step 4: Predict the Cost for 1500 miles

Substitute \( d = 1500 \) into the equation \( C = \frac{1}{4}d + 260 \): \[C = \frac{1}{4}(1500) + 260 = 375 + 260 = 635.\] The predicted cost for driving 1500 miles is \( \$635 \).

Step 5: Graph the Linear Equation

The linear graph of \( C = \frac{1}{4}d + 260 \) is a straight line. The slope \( \frac{1}{4} \) represents the variable cost per mile. The \( y \)-intercept \( 260 \) represents the fixed monthly cost when no miles are driven.

Step 6: Interpreting the Slope and Y-intercept

The slope \( \frac{1}{4} \) signifies that for every additional mile driven, the cost increases by \( \$0.25 \). The \( y \)-intercept \( 260 \) symbolizes the fixed cost component, such as insurance or taxes, which does not depend on distance driven.

Step 7: Justify the Linear Relationship

A linear model is suitable here because the costs appear to increase uniformly with distance, implying constant variable costs per mile.

Key Concepts

Slope of a Line

The slope of a line is a crucial concept in understanding linear equations. In this particular exercise, the slope represents how the monthly cost of driving changes concerning the distance driven. We calculate the slope, denoted as \( m \), using the slope formula:

• \( m = \frac{C_2 - C_1}{d_2 - d_1} \)

This formula essentially measures the change in the cost \( C \) for a change in distance \( d \). Here, the slope was found to be \( \frac{1}{4} \), meaning the cost increases by \$0.25 for every mile driven. This is an example of a constant rate of change, something essential in understanding linear relationships.

Y-intercept

The y-intercept of a linear equation is where the line crosses the y-axis. In other words, it is the value of \( C \) when \( d = 0 \). In our exercise, the y-intercept is 260, as calculated when solving the linear equation. The y-intercept illustrates the fixed cost, which applies even when no miles are driven. This could include costs such as car insurance or lease payments, which stay constant regardless of driving distance. Understanding the y-intercept helps us see the complete picture of monthly car expenses.

Linear Relationship

A linear relationship exists when two variables have a constant rate of change. In this case, the monthly cost \( C \) and distance driven \( d \) share a linear relationship, as suggested by the exercise. The line equation \( C = \frac{1}{4}d + 260 \) derived from the problem confirms that the cost increases by a fixed amount per mile.

• Consistency: This relationship implies predictable changes in cost with changes in distance driven.
• Graph Representation: On a graph, this relationship appears as a straight line, indicating steadiness.

Understanding linear relationships helps in making predictions and understanding trends in practical situations like driving costs.

Cost Prediction

Cost prediction using a linear equation allows one to estimate future costs based on known variables. In this exercise, we predict the cost of driving 1500 miles using the linear equation obtained:

• \( C = \frac{1}{4}(1500) + 260 \)
• This yields a prediction of \$635.

Predicting costs is a valuable tool in budgeting and planning. It underscores how linear equations can be applied to everyday problems, providing a useful method for anticipating future expenses based on current patterns.