91Ó°ÊÓ

A sports trainer has monthly costs of \(\$ 69.95\) for phone service and \(\$ 39.99\) for his website and advertising. In addition he pays a \(\$ 20\) fee to the gym for each session in which he trains a client. a. Write a cost function to represent the \(\operatorname{cost} C(x)\) for \(x\) training sessions. b. Write a function representing the average \(\operatorname{cost} \bar{C}(x)\) for \(x\) sessions. c. Evaluate \(\bar{C}(5), \bar{C}(30),\) and \(\bar{C}(120)\). d. The trainer can realistically have 120 sessions per month. However, if the number of sessions were unlimited, what value would the average cost approach? What does this mean in the context of the problem?

Step by step solution

01

Define the Monthly Fixed Costs

Identify the fixed costs: phone service (\(69.95) and website & advertising (\)39.99). Add these to get the total fixed costs. Fixed costs = 69.95 + 39.99 = 109.94

02

Define the Variable Cost per Session

Identify the variable cost: fee paid to the gym per session ($20). Total variable cost for x sessions is 20x.

03

Write the Cost Function

Combine the fixed and variable costs into one cost function.Cost function: C(x) = 109.94 + 20x

04

Write the Average Cost Function

To find the average cost function, divide the total cost function by the number of sessions, x.Average cost function: \(\bar{C}(x) = \frac{109.94 + 20x}{x}\)

05

Evaluate \bar{C}(5), \bar{C}(30), and \bar{C}(120)

Substitute x with 5, 30, and 120 in the average cost function and simplify.\(\bar{C}(5) = \frac{109.94 + 20(5)}{5} = \frac{209.94}{5} = 41.99\)\(\bar{C}(30) = \frac{109.94 + 20(30)}{30} = \frac{709.94}{30} = 23.665\)\(\bar{C}(120) = \frac{109.94 + 20(120)}{120} = \frac{2509.94}{120} = 20.91617\)

06

Determine the Average Cost as x Approaches Infinity

Analyze the function \(\bar{C}(x) = \frac{109.94 + 20x}{x}\). As the number of sessions, x, increases indefinitely, the average cost function approaches 20. This is because the fixed cost (109.94) becomes insignificant compared to the total variable cost (20x). So \(\lim_{{x \to \infty}} \bar{C}(x) = 20\)

07

Contextual Meaning

When the number of sessions is unlimited, the average cost per session stabilizes at 20, indicating this is the per-session fee paid to the gym. This means, practically, the fixed monthly costs are spread so thin over a large number of sessions that they hardly impact the average cost.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

fixed costs

Fixed costs are the expenses that remain constant regardless of the number of services or goods produced. In the scenario provided, the sports trainer has set monthly costs for phone service and website advertising. These are considered fixed because they do not change with the number of training sessions conducted.
To calculate these fixed costs, we simply add them up:
\( 69.95 + 39.99 = 109.94 \).
So, the trainer has fixed costs of \( \$ 109.94 \) every month. Even if the trainer has 0 or 100 sessions, this expense does not change.

variable costs

Variable costs change directly with the production volume. In this example, it's the fee the trainer pays the gym for each session. The variable cost per session is \( \$ 20 \).
To find the total variable cost for any number of sessions, let's denote it as \( x \):
\( 20x \).
So, if the trainer conducts 5 sessions, the variable cost will be \( 20 \times 5 = 100 \). Similarly, for 30 or 120 sessions, you would multiply 20 by 30 or 120, respectively.
This means the total cost depends on how many sessions the trainer conducts.

average cost function

The average cost function helps us understand how costs distribute per unit of production, or in this case, per training session. To find the average cost per session, we divide the total cost by the number of sessions.
Given the cost function \( C(x) = 109.94 + 20x \):
Average cost function \( \bar{C}(x) = \frac{109.94 + 20x}{x} \).
This function simplifies to understand how much each session costs on average. For instance, with 5 sessions:
\( \bar{C}(5) = \frac{109.94 + 20 \times 5}{5} = \frac{209.94}{5} = 41.99 \). Doing this for other values, like 30 or 120 sessions, follows the same method.

limit of a function

The concept of the limit of a function helps us understand the behavior of a function as the input grows very large. For the average cost function \( \bar{C}(x) = \frac{109.94 + 20x}{x} \):
As \( x \) approaches infinity, the term \( \frac{109.94}{x} \) becomes very small and tends to zero. Therefore:
\( \lim_{{x \to \infty}} \bar{C}(x) = \lim_{{x \to \infty}} \frac{109.94}{x} + 20 = 20 \).
Practically, this means that as the trainer conducts more and more sessions, the impact of the fixed costs diminishes, and the average cost per session will settle at \( \$ 20 \).