A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by C=0.3x² +9,000 How fast is the cost per month changing when production is changing at the rate of 11 units per month and the production level is 80 units? at the rate of $ per month at this production level Costs are (Round to eded) decreasing increasing
A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by C=0.3x² +9,000 How fast is the cost per month changing when production is changing at the rate of 11 units per month and the production level is 80 units? at the rate of $ per month at this production level Costs are (Round to eded) decreasing increasing
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter5: Exponential And Logarithmic Functions
Section5.2: Applications Of Exponential Functions
Problem 43E
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![### Cost Analysis of Wine Rack Production
A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by the following cost function:
\[ C = 0.3x^2 + 9000 \]
#### Problem Statement
Determine how fast the cost per month is changing when production is increasing at a rate of 11 units per month and the production level is 80 units.
To understand and solve this problem, follow these steps.
1. **Cost Function Derivative**: Differentiate the given cost function \( C \) with respect to time \( t \).
2. **Determine the Rate of Change**: Use the given production rate and production level in your calculations.
#### Derivative Calculation
Given:
\[ C(x) = 0.3x^2 + 9000 \]
Differentiate \( C \) with respect to \( x \):
\[ \frac{dC}{dx} = 0.6x \]
#### Applying the Given Rates
Given:
Production rate \( \frac{dx}{dt} = 11 \) units/month
Production level \( x = 80 \)
Use the chain rule to find \( \frac{dC}{dt} \):
\[ \frac{dC}{dt} = \frac{dC}{dx} \cdot \frac{dx}{dt} \]
Substitute the values into the derivative:
\[ \frac{dC}{dt} = 0.6 \cdot 80 \cdot 11 \]
\[ \frac{dC}{dt} = 528 \]
Thus, the cost per month is changing at a rate of 528 monetary units when the production level is 80 units and increasing at a rate of 11 units per month.
#### Answer Selection
In the given dropdown options and blank fields:
- Fill in "increasing".
- The rate of increase is "528".
- The production level is "80".
Ensure to round off the answer to the nearest monetary unit if required.
##### Detailed Diagram Explanation
Since there are no graphs or diagrams in the image, no further visual explanation is needed.
This interactive exercise helps students practice derivative applications in economics and cost analysis. Fill in the answers in the provided fields, select the appropriate option stating the cost change (increasing), and submit to verify your solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F02b4a1f4-9e35-42db-aa2a-674bb26df94b%2F54377902-1bd3-4ae2-b6a5-789ea6c55e5a%2Fl8uxyk9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Cost Analysis of Wine Rack Production
A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by the following cost function:
\[ C = 0.3x^2 + 9000 \]
#### Problem Statement
Determine how fast the cost per month is changing when production is increasing at a rate of 11 units per month and the production level is 80 units.
To understand and solve this problem, follow these steps.
1. **Cost Function Derivative**: Differentiate the given cost function \( C \) with respect to time \( t \).
2. **Determine the Rate of Change**: Use the given production rate and production level in your calculations.
#### Derivative Calculation
Given:
\[ C(x) = 0.3x^2 + 9000 \]
Differentiate \( C \) with respect to \( x \):
\[ \frac{dC}{dx} = 0.6x \]
#### Applying the Given Rates
Given:
Production rate \( \frac{dx}{dt} = 11 \) units/month
Production level \( x = 80 \)
Use the chain rule to find \( \frac{dC}{dt} \):
\[ \frac{dC}{dt} = \frac{dC}{dx} \cdot \frac{dx}{dt} \]
Substitute the values into the derivative:
\[ \frac{dC}{dt} = 0.6 \cdot 80 \cdot 11 \]
\[ \frac{dC}{dt} = 528 \]
Thus, the cost per month is changing at a rate of 528 monetary units when the production level is 80 units and increasing at a rate of 11 units per month.
#### Answer Selection
In the given dropdown options and blank fields:
- Fill in "increasing".
- The rate of increase is "528".
- The production level is "80".
Ensure to round off the answer to the nearest monetary unit if required.
##### Detailed Diagram Explanation
Since there are no graphs or diagrams in the image, no further visual explanation is needed.
This interactive exercise helps students practice derivative applications in economics and cost analysis. Fill in the answers in the provided fields, select the appropriate option stating the cost change (increasing), and submit to verify your solution.
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