Consider the following cost function. a. Find the average cost and marginal cost functions. b. Determine the average and marginal cost when x = a. c. Interpret the values obtained in part (b). C(x)=1600+0.1x, 0≤x≤ 5000, a = 2900 a. The average cost function is C(x) = SHE
Consider the following cost function. a. Find the average cost and marginal cost functions. b. Determine the average and marginal cost when x = a. c. Interpret the values obtained in part (b). C(x)=1600+0.1x, 0≤x≤ 5000, a = 2900 a. The average cost function is C(x) = SHE
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Parts a,b,c
![**Cost Function Analysis**
Consider the following cost function:
\[C(x) = 1600 + 0.1x, \quad 0 \leq x \leq 5000, \quad a = 2900\]
**a. Find the average cost and marginal cost functions.**
The average cost function is given by \(\bar{C}(x)\).
**b. Determine the average and marginal cost when \(x = a\).**
**c. Interpret the values obtained in part (b).**
### Solution:
Let's find the required functions and values.
1. **Average Cost Function:**
The average cost function \(\bar{C}(x)\) is calculated by dividing the total cost function \(C(x)\) by the quantity \(x\):
\[
\bar{C}(x) = \frac{C(x)}{x} = \frac{1600 + 0.1x}{x}
\]
Simplifying the expression:
\[
\bar{C}(x) = \frac{1600}{x} + 0.1
\]
2. **Marginal Cost Function:**
The marginal cost function is the derivative of the total cost function \(C(x)\) with respect to \(x\):
\[
C'(x) = \frac{d}{dx}(1600 + 0.1x)
\]
\[
C'(x) = 0 + 0.1 = 0.1
\]
3. **Average and Marginal Cost at \(x = a = 2900\):**
- **Average Cost:**
\[
\bar{C}(2900) = \frac{1600}{2900} + 0.1 = \frac{1600}{2900} + 0.1
\]
Simplify:
\[
\bar{C}(2900) \approx 0.5517
\]
- **Marginal Cost:**
The marginal cost is constant for all \(x\), therefore:
\[
C'(2900) = 0.1
\]
4. **Interpretation:**
- The average cost at \(x = 2900\) units is approximately 0.5517.
- The marginal cost is constant](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa589a0b1-5940-47b0-b798-8fa24b0ff23c%2F27f08b93-0fbb-4076-b7df-107f4a2fafe2%2Fxs4o78_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Cost Function Analysis**
Consider the following cost function:
\[C(x) = 1600 + 0.1x, \quad 0 \leq x \leq 5000, \quad a = 2900\]
**a. Find the average cost and marginal cost functions.**
The average cost function is given by \(\bar{C}(x)\).
**b. Determine the average and marginal cost when \(x = a\).**
**c. Interpret the values obtained in part (b).**
### Solution:
Let's find the required functions and values.
1. **Average Cost Function:**
The average cost function \(\bar{C}(x)\) is calculated by dividing the total cost function \(C(x)\) by the quantity \(x\):
\[
\bar{C}(x) = \frac{C(x)}{x} = \frac{1600 + 0.1x}{x}
\]
Simplifying the expression:
\[
\bar{C}(x) = \frac{1600}{x} + 0.1
\]
2. **Marginal Cost Function:**
The marginal cost function is the derivative of the total cost function \(C(x)\) with respect to \(x\):
\[
C'(x) = \frac{d}{dx}(1600 + 0.1x)
\]
\[
C'(x) = 0 + 0.1 = 0.1
\]
3. **Average and Marginal Cost at \(x = a = 2900\):**
- **Average Cost:**
\[
\bar{C}(2900) = \frac{1600}{2900} + 0.1 = \frac{1600}{2900} + 0.1
\]
Simplify:
\[
\bar{C}(2900) \approx 0.5517
\]
- **Marginal Cost:**
The marginal cost is constant for all \(x\), therefore:
\[
C'(2900) = 0.1
\]
4. **Interpretation:**
- The average cost at \(x = 2900\) units is approximately 0.5517.
- The marginal cost is constant
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