Problem 97 - Intermediate Algebra

Question

An electric company charges \(\$ 7.50\) per month plus \(9 \notin\) for each kilowatt hour ( \(\mathrm{kwh}\) ) of electricity used. a. Find a linear function that gives the total cost of \(n\) kwh of electricity. (Hint: See Example 1.) b. Find a rational function that gives the average cost per \(\mathrm{kwh}\) when using \(n\) kwh. c. Find the average cost per kwh when 775 kwh are used.

Step-by-Step Solution

Answer

a. \(C(n) = 7.50 + 0.09n\); b. \(A(n) = \frac{7.50}{n} + 0.09\); c. \$0.0997 per kwh.

Step 1: Define the Linear Cost Function

The cost function is composed of a fixed fee and a variable fee based on electricity usage. The fixed cost is \(7.50 per month, and the variable cost is \)0.09 (or 9 cents) per kwh. Therefore, the linear function for the total cost in terms of the number of kwh, \(n\), is: \( C(n) = 7.50 + 0.09n \).

Step 2: Rational Function for Average Cost

The average cost function is found by dividing the total cost by the number of kwh used, \(n\). Thus, the rational function for the average cost per kwh is: \( A(n) = \frac{C(n)}{n} = \frac{7.50 + 0.09n}{n} = \frac{7.50}{n} + 0.09 \).

Step 3: Calculate Average Cost for 775 kwh

Using the formula \( A(n) = \frac{7.50}{n} + 0.09 \), substitute \(n = 775\) to find the average cost. This gives: \[ A(775) = \frac{7.50}{775} + 0.09 \approx 0.0997 \]. So, the average cost per kwh when 775 kwh are used is approximately \$0.0997, or 9.97 cents per kwh.

Key Concepts

Rational Functions and Their Role in Cost Calculation

Rational functions are mathematical expressions that involve ratios of polynomials. In the context of electricity usage, a rational function can be very handy for determining costs that vary with a dependent variable—in this case, the kilowatt-hours (kwh) used.
When we're calculating the cost of electricity, it is helpful to think about how the cost changes with each unit of electricity consumed. This can be captured effectively using rational functions. In step 2 of our original solution, we see the formation of the rational function to express average cost per kwh:

• The total cost (the numerator) is divided by the number of units (kwh used), leading to the average cost function, \( A(n) = \frac{7.50 + 0.09n}{n} \).
• This formula simplifies to \( A(n) = \frac{7.50}{n} + 0.09 \), showing both a constant cost per unit and a portion that decreases as the usage becomes higher, represented by \( \frac{7.50}{n} \).

The rational function is essential here because it helps us comprehend and quantify how average cost behaves as usage changes. With such a function, it's easier to predict expenses over time and make informed decisions about consumption.

Understanding Average Cost Calculation

To grasp how average cost calculations work, imagine you're trying to find out how much, on average, each unit of electricity costs. This calculation becomes particularly insightful when costs include both fixed and variable components. Let's break it down:

• Firstly, you compute the total cost, which includes both a fixed part, \( \\(7.50 \) in our case, and a variable part that accounts for actual usage at \( \\)0.09 \) per kwh.
• The average cost per kwh is derived by spreading out the total cost over all the units (kwh) used. This is mathematically expressed as \( A(n) = \frac{7.50 + 0.09n}{n} \).
• On simplifying the expression, the result \( \frac{7.50}{n} + 0.09 \) indicates that the regular cost of electricity per each additional kwh is \( \$0.09 \), plus a variable fraction determined by \( \frac{7.50}{n} \).

The key insight from average cost calculations is that as you use more electricity, the influence of the fixed cost becomes less significant per unit, reducing the overall average cost with rising usage.

Modeling Electricity Usage Cost

Modeling the cost of electricity usage using linear and rational functions is instrumental for cost prediction and management. Such modeling helps households and businesses understand how different usage levels will affect their bill.
The linear function \( C(n) = 7.50 + 0.09n \) gives the total cost for any number of used kwh, \( n \). It combines

• a fixed monthly fee, ensuring there's a base charge regardless of consumption, and
• a variable charge, which scales with the amount of electricity used.

Additionally, rational functions allow us to see how the **average** cost per kwh changes with different levels of usage. When you calculate the average cost for a specific amount of kwh, say 775, using the formula \( A(n) = \frac{7.50}{n} + 0.09 \), you get a precise figure for that usage, which in our example is approximately \( \$0.0997 \) or 9.97 cents per kwh.
This modeling is practical as it not only prepares you for budgeting but also offers insights for optimizing electricity usage to minimize expenses, especially when faced with both fixed and variable billing components.