Problem 6 - Precalculus Essentials

Question

A discount pass for a bridge costs \(\$ 21\) per month. The toll for the bridge is normally \(\$ 2.50\), but it is reduced to \(\$ 1\) for people who have purchased the discount pass. a. Express the total monthly cost to use the bridge without a discount pass, \(f,\) as a function of the number of times in a month the bridge is crossed, \(x\). b. Express the total monthly cost to use the bridge with a discount pass, \(g,\) as a function of the number of times in a month the bridge is crossed, \(x\). c. Determine the number of times in a month the bridge must be crossed so that the total monthly cost without the discount pass is the same as the total monthly cost with the discount pass. What will be the monthly cost for each option?

Step-by-Step Solution

Answer

With or without the discount pass, the total monthly cost to use the bridge is \$35 when the bridge is crossed exactly 14 times in a month.

Step 1: Define function f(x) for the cost without the pass

This function is linear with a slope equal to the cost per crossing, or \(2.50\). There's no monthly constant fee. Therefore, \(f(x) = 2.5x\).

Step 2: Define function g(x) for the cost with the discount pass

This function is also linear but includes a monthly constant of \(21\) representing the cost of the pass. Each crossing costs \(1\). Therefore, \(g(x) = 21 + x\).

Step 3: Determine the point of intersection

The costs become equal when \(f(x) = g(x)\). That gives us the equation \(2.5x = 21 + x\). Re-arrange it to find \(x\).

Step 4: Solve the equation for x

Isolate \(x\) to one side by subtracting \(x\) from both sides, yielding \(2.5x - x = 21\), which simplifies to \(1.5x = 21\). Divide both sides by \(1.5\) to find the value of \(x\), giving \(x = 14\).

Step 5: Find the monthly cost for both options at this number of crossings

Substitute \(x = 14\) into either function, for instance \(g(x) = 21 + 14 = 35\). So, the cost is \$35 for both options when the bridge is crossed 14 times in a month.

Key Concepts

Understanding Linear Functions in Cost Analysis

Linear functions are mathematical expressions that represent relationships with a constant rate of change. They are useful for modeling situations where one quantity depends on another linearly, meaning the change occurs in consistent steps. In cost analysis, linear functions can simplify understanding how costs increase with usage.
For our bridge toll problem, we have two linear functions: the cost without the pass, and the cost with the pass. The function without the pass is expressed as \( f(x) = 2.5x \). Here, \( f(x) \) represents the total cost in dollars, and \( x \) is the number of crossings. This shows that the cost increases by \$2.50 for each crossing.
On the other hand, the function with the discount pass includes a constant monthly fee, represented as \( g(x) = 21 + x \). The number \( 21 \) is the cost of buying the pass per month, while \( x \), the slope of 1, indicates the cost increase per crossing at a reduced rate. Understanding these functions helps us compare total costs based on crossing frequency.

Making a Cost Comparison

Cost comparison in this scenario involves assessing which option offers the better value based on the number of bridge crossings. This process helps to pinpoint the exact conditions where costs with and without the discount pass are equal.
The first option, without a discount pass, costs \( 2.50 \) per crossing. Conversely, the second option introduces a fixed cost of \\(21 plus \\)1 per crossing. Both options are linear but exhibit different rates of change.

• Without the pass: The cost scales by \\(2.50 per crossing, making it straightforward from the first crossing at zero initial cost.
• With the pass: The total cost starts higher due to the \\)21 pass fee but increases slowly at \$1 per crossing.

The key to effective cost comparison is finding where these two costs intersect — that is, the point at which the cumulative cost for both options is the same. This involves solving the functions to determine this crossover point.

Solving Equations to Find the Cost Equal Point

Equation solving is pivotal in determining when the costs of two options match. For our bridge example, we set the two cost functions equal to find the crossover point, where using a discount pass becomes cost-effective.
We equate the two linear functions: \( f(x) = g(x) \). This translates to the equation \( 2.5x = 21 + x \). Solving for \( x \) helps us pinpoint the precise number of crossings where costs are equivalent.

• First, subtract \( x \) from both sides: \( 2.5x - x = 21 \).
• Simplify it to \( 1.5x = 21 \).

Finally, divide both sides by \( 1.5 \) to find \( x = 14 \). This tells us that crossing the bridge 14 times in a month equalizes the monthly costs, resulting in \$35 for both options. This solution highlights how solving equations can determine cost patterns and optimize financial decisions.