The tolal Gost of produging × Shirts is: C(x)= Gom +30x Find the rate gt which this qverage. Gast is Ghanging when 10 shirts gra boing produded. bei
The tolal Gost of produging × Shirts is: C(x)= Gom +30x Find the rate gt which this qverage. Gast is Ghanging when 10 shirts gra boing produded. bei
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Understanding the Rate of Change in Average Cost
The following problem explores the concept of how costs change as production levels vary.
**Problem:**
The total cost of producing \( x \) shirts is given by the function \( C(x) = 6000 + 30x \).
**Objective:**
Find the rate at which the average cost is changing when 10 shirts are being produced.
---
**Explanation:**
1. **Total Cost Function:**
\[
C(x) = 6000 + 30x
\]
2. **Average Cost (AC):**
\[
AC(x) = \frac{C(x)}{x} = \frac{6000 + 30x}{x} = \frac{6000}{x} + 30
\]
3. **Rate of Change of Average Cost:**
To find how the average cost changes with respect to the number of shirts produced, we calculate the derivative of \( AC(x) \) with respect to \( x \).
\[
AC'(x) = \frac{d}{dx}\left( \frac{6000}{x} + 30 \right)
\]
Simplifying,
\[
AC'(x) = \frac{d}{dx} \left( \frac{6000}{x} \right)
\]
\[
AC'(x) = 6000 \cdot \left( \frac{d}{dx} (x^{-1}) \right)
\]
\[
AC'(x) = 6000 \cdot (-x^{-2})
\]
\[
AC'(x) = - \frac{6000}{x^2}
\]
4. **Calculating When \( x = 10 \):**
\[
AC'(10) = - \frac{6000}{10^2} = - \frac{6000}{100} = -60
\]
**Conclusion:**
The rate at which the average cost is changing when 10 shirts are being produced is -60. This means that for each additional shirt produced, the average cost decreases by 60 units.
This exercise helps in understanding the relationship between production volume and cost efficiency, which is crucial in manufacturing and economics.
### Graphical Interpretation (if applicable):
- If a graph was provided, it might](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb93a4b56-e13a-4118-8823-601610d155f6%2Fd40cea79-6681-4200-b8f4-57ba759f44db%2F2vo8m0k_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding the Rate of Change in Average Cost
The following problem explores the concept of how costs change as production levels vary.
**Problem:**
The total cost of producing \( x \) shirts is given by the function \( C(x) = 6000 + 30x \).
**Objective:**
Find the rate at which the average cost is changing when 10 shirts are being produced.
---
**Explanation:**
1. **Total Cost Function:**
\[
C(x) = 6000 + 30x
\]
2. **Average Cost (AC):**
\[
AC(x) = \frac{C(x)}{x} = \frac{6000 + 30x}{x} = \frac{6000}{x} + 30
\]
3. **Rate of Change of Average Cost:**
To find how the average cost changes with respect to the number of shirts produced, we calculate the derivative of \( AC(x) \) with respect to \( x \).
\[
AC'(x) = \frac{d}{dx}\left( \frac{6000}{x} + 30 \right)
\]
Simplifying,
\[
AC'(x) = \frac{d}{dx} \left( \frac{6000}{x} \right)
\]
\[
AC'(x) = 6000 \cdot \left( \frac{d}{dx} (x^{-1}) \right)
\]
\[
AC'(x) = 6000 \cdot (-x^{-2})
\]
\[
AC'(x) = - \frac{6000}{x^2}
\]
4. **Calculating When \( x = 10 \):**
\[
AC'(10) = - \frac{6000}{10^2} = - \frac{6000}{100} = -60
\]
**Conclusion:**
The rate at which the average cost is changing when 10 shirts are being produced is -60. This means that for each additional shirt produced, the average cost decreases by 60 units.
This exercise helps in understanding the relationship between production volume and cost efficiency, which is crucial in manufacturing and economics.
### Graphical Interpretation (if applicable):
- If a graph was provided, it might
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