Problem 96 - Intermediate Algebra

Question

The average (mean) cost for a service club to publish a directory of its members is given by the rational function $$ f(x)=\frac{1.25 x+700}{x} $$ where \(x\) is the number of directories printed. Find the average cost per directory if a. 500 directories are printed. b. \(2,000\) directories are printed.

Step-by-Step Solution

Answer

a. $2.65; b. $1.60.

Step 1: Understand the Problem

We are given a rational function that represents the average cost of printing directories, where the number of directories printed is represented by \(x\). Our job is to find the average cost when printing 500 and 2,000 directories.

Step 2: Substitute for 500 Directories

To find the average cost when 500 directories are printed, substitute \(x = 500\) into the function: \[f(500) = \frac{1.25(500) + 700}{500}\]

Step 3: Calculate for 500 Directories

Multiply 1.25 by 500 and then add 700:\[1.25 \times 500 = 625\]\[625 + 700 = 1325\]Now divide by 500 to find average cost:\[f(500) = \frac{1325}{500}\]

Step 4: Simplify for 500 Directories

Simplify the fraction:\[f(500) = 2.65\]This means the average cost for printing 500 directories is $2.65.

Step 5: Substitute for 2000 Directories

Next, find the average cost when 2,000 directories are printed by substituting \(x = 2000\) into the function:\[f(2000) = \frac{1.25(2000) + 700}{2000}\]

Step 6: Calculate for 2000 Directories

Multiply 1.25 by 2000 and then add 700:\[1.25 \times 2000 = 2500\]\[2500 + 700 = 3200\]Now, divide by 2000 to find average cost:\[f(2000) = \frac{3200}{2000}\]

Step 7: Simplify for 2000 Directories

Simplify the fraction:\[f(2000) = 1.60\]This means the average cost for printing 2,000 directories is $1.60.

Key Concepts

Average Cost Calculation

When dealing with average cost calculation, we often use rational functions to relate total cost and the number of items produced. In this context, the function \( f(x) = \frac{1.25x + 700}{x} \) represents the average cost per directory. Here, \( x \) is the number of directories printed. To find the average cost, we substitute \( x \) with specific values. This allows us to calculate the average cost based on different production scales.
For example, in the exercise, we calculate the average cost by substituting 500 and 2,000 for \( x \). By performing these calculations, we find that more directories result in a lower average cost per directory. This demonstrates economies of scale in production, where costs decrease as volume increases.

Substitution Method

The substitution method simplifies finding specific values of a function for given inputs. By substituting numbers, we can evaluate the function at these particular points. In our case, we are finding the average cost by plugging in specific values of \( x \), the number of directories printed, into the function.

• For 500 directories, replace \( x \) in \( f(x) = \frac{1.25x + 700}{x} \) with 500. This allows us to compute \( f(500) = \frac{1.25(500) + 700}{500} \).
• Similarly, for 2,000 directories, substitute \( x = 2000 \) to get \( f(2000) = \frac{1.25(2000) + 700}{2000} \).

This approach helps to analyze how the average cost changes with different production levels, providing useful insights into cost management.

Simplifying Fractions

Simplifying fractions is an essential arithmetic skill, allowing us to reduce fractions to their simplest form for easier interpretation and comparison. In rational functions, simplification helps convey the function's output clearly.
For example, after calculating the numerator in the average cost function, the fraction may appear complex. By dividing both the numerator and the denominator by their greatest common divisor, we simplify:

• In step 4, after calculations, we get \( \frac{1325}{500} \). Simplifying gives us the result \( 2.65 \).
• In step 7, the fraction \( \frac{3200}{2000} \) simplifies to \( 1.60 \).

These simpler forms make interpreting the results easier, particularly when comparing average costs. Understanding how to simplify fractions enables us to present and analyze mathematical findings more clearly.