A coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is represented by the function below, where x is the number of pounds of coffee roasted. Find the total cost of roasting 470 lb of coffee, disregarding any fixed costs. C'(x) = -0.014x+8.25, for x ≤ 500 The total cost is $ (Round to the nearest cent as needed.)
A coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is represented by the function below, where x is the number of pounds of coffee roasted. Find the total cost of roasting 470 lb of coffee, disregarding any fixed costs. C'(x) = -0.014x+8.25, for x ≤ 500 The total cost is $ (Round to the nearest cent as needed.)
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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![### Calculating the Total Cost of Coffee Roasting
A coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is represented by the function below, where \( x \) is the number of pounds of coffee roasted. Find the total cost of roasting 470 lb of coffee, disregarding any fixed costs.
\[ C'(x) = -0.014x + 8.25, \quad \text{for} \quad x \leq 500 \]
---
### Problem Solution
The total cost is \( \$ \) [ ].
*(Round to the nearest cent as needed.)*
### Explanation
Given a marginal cost function \( C'(x) \), to find the total cost of roasting \( x \) pounds of coffee, we need to integrate the marginal cost function:
\[ C(x) = \int C'(x) \, dx \]
The given function is:
\[ C'(x) = -0.014x + 8.25 \]
Integrate this function with respect to \( x \):
\[ C(x) = \int (-0.014x + 8.25) \, dx \]
This results in:
\[ C(x) = -0.007x^2 + 8.25x + C \]
Where \( C \) is the constant of integration. Since we are disregarding fixed costs, assume \( C \) is zero for simplicity.
Thus, the total cost function becomes:
\[ C(x) = -0.007x^2 + 8.25x \]
To find the total cost for roasting 470 pounds of coffee, substitute \( x = 470 \) into the total cost function:
\[ C(470) = -0.007(470)^2 + 8.25(470) \]
Calculating \( 470^2 \):
\[ 470^2 = 220900 \]
Substitute this back into the equation:
\[ C(470) = -0.007(220900) + 8.25(470) \]
\[ C(470) = -1546.3 + 3877.5 \]
\[ C(470) = 2331.2 \]
Hence, the total cost of roasting 470 pounds of coffee is:
\[ \boxed{2331.20} \]
---
### Conclusion
By integrating the marginal cost function and evaluating it at](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febbbc1cc-ba98-4671-a28a-276fb59e8560%2F9fd7f041-2b64-4222-8e4f-386aaab6b3a3%2F565jrv7_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Total Cost of Coffee Roasting
A coffee company has found that the marginal cost, in dollars per pound, of the coffee it roasts is represented by the function below, where \( x \) is the number of pounds of coffee roasted. Find the total cost of roasting 470 lb of coffee, disregarding any fixed costs.
\[ C'(x) = -0.014x + 8.25, \quad \text{for} \quad x \leq 500 \]
---
### Problem Solution
The total cost is \( \$ \) [ ].
*(Round to the nearest cent as needed.)*
### Explanation
Given a marginal cost function \( C'(x) \), to find the total cost of roasting \( x \) pounds of coffee, we need to integrate the marginal cost function:
\[ C(x) = \int C'(x) \, dx \]
The given function is:
\[ C'(x) = -0.014x + 8.25 \]
Integrate this function with respect to \( x \):
\[ C(x) = \int (-0.014x + 8.25) \, dx \]
This results in:
\[ C(x) = -0.007x^2 + 8.25x + C \]
Where \( C \) is the constant of integration. Since we are disregarding fixed costs, assume \( C \) is zero for simplicity.
Thus, the total cost function becomes:
\[ C(x) = -0.007x^2 + 8.25x \]
To find the total cost for roasting 470 pounds of coffee, substitute \( x = 470 \) into the total cost function:
\[ C(470) = -0.007(470)^2 + 8.25(470) \]
Calculating \( 470^2 \):
\[ 470^2 = 220900 \]
Substitute this back into the equation:
\[ C(470) = -0.007(220900) + 8.25(470) \]
\[ C(470) = -1546.3 + 3877.5 \]
\[ C(470) = 2331.2 \]
Hence, the total cost of roasting 470 pounds of coffee is:
\[ \boxed{2331.20} \]
---
### Conclusion
By integrating the marginal cost function and evaluating it at
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