Question

Truck Rentals A truck rental company rents a moving truck for one day by charging \(\$ 39\) plus \(\$ 0.60\) per mile. Write a linear equation that relates the cost \(C,\) in dollars, of renting the truck to the number \(x\) of miles driven. What is the cost of renting the truck if the truck is driven 110 miles? 230 miles?

Short answer

The linear equation is \(C = 39 + 0.60x\). The cost for 110 miles is \(\$105\), and for 230 miles, it is \(\$177\).

Step-by-step solution

Identify the Variable

Define the variable. Let the number of miles driven be represented by \(x\).

Define the Constants

Determine the constants given in the problem. The base cost of renting the truck is \(\$39\) and the additional cost per mile is \(\$0.60\).

Write the Linear Equation

Formulate the linear equation: The total cost \(C\) is the sum of the base cost and the product of the cost per mile and the number of miles driven. Therefore, the equation is \[C = 39 + 0.60x\]

Calculate the Cost for 110 Miles

Substitute \(x = 110\) into the equation and solve for \(C\): \[C = 39 + 0.60(110) = 39 + 66 = 105\] Therefore, the cost for 110 miles is \(\$105\).

Calculate the Cost for 230 Miles

Substitute \(x = 230\) into the equation and solve for \(C\): \[C = 39 + 0.60(230) = 39 + 138 = 177\] Therefore, the cost for 230 miles is \(\$177\).

Key concepts

Cost Function

A cost function is a mathematical formula used to calculate the total cost of a service or product, given a set of variables. In this exercise, we are dealing with the cost function associated with renting a truck. The formula we derived is: \[ C = 39 + 0.60x \] Here, 'C' represents the total cost of renting the truck, and 'x' is the number of miles driven.
The base cost of \$39\ remains constant, no matter how many miles you drive. This makes it the \constant term\ in our equation. The term \$0.60x\ varies depending on the number of miles, making it the \variable term\. The cost function helps us easily calculate what the total cost will be, using a simple substitution of the number of miles into the equation.
By understanding cost functions, we can better manage expenses and predict future costs based on different scenarios.

Linear Relationships

Linear relationships are a core concept in algebra where two variables have a straight-line relationship. For example, in this truck rental problem, we found that the cost (C) depends linearly on the number of miles driven (x).
This relationship can be graphed as a straight line where the y-axis represents the total cost, and the x-axis represents the miles driven. The formula \[ C = 39 + 0.60x \] is in the form of \y = mx + b\, where 'm' is the slope (0.60 in our case) and 'b' represents the y-intercept (39 in our case).
The 'slope' indicates how steep the line is, or in this scenario, how much the cost increases per mile driven. The 'intercept' tells us the initial cost when no miles have been driven. Linear relationships make it easy to visualize and understand how two quantities are related and how changes in one affect the other.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operators (like + or -) that represent mathematical relationships. In our problem, we used an algebraic expression to build the cost function:\[ C = 39 + 0.60x \] Let's break it down:

• \textbf{39:} This is a constant, representing a fixed cost.
• \textbf{0.60:} This is a coefficient, representing cost per mile.
• \textbf{x:} This is a variable, representing the number of miles driven.

When \solving algebraic expressions,\ it's important to follow the order of operations and correctly substitute variables with their respective values.
Using these basic principles of algebra, we can handle more complex equations and start solving real-world problems. Always remember, practice makes perfect, and breaking down each part helps in understanding the entire expression.