nd price (demand) functions C(q) = 100q+43,700 and p(q) = 2q + 890, what is the marginal revenue at a production level of revenue is dollars per item. er to nearest dollar.)
nd price (demand) functions C(q) = 100q+43,700 and p(q) = 2q + 890, what is the marginal revenue at a production level of revenue is dollars per item. er to nearest dollar.)
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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![**Problem:**
Given cost and price (demand) functions \( C(q) = 100q + 43,700 \) and \( p(q) = -2q + 890 \), what is the marginal revenue at a production level of 60 items?
The marginal revenue is \(\_\_\_\_\) dollars per item.
(Round answer to nearest dollar.)
**Answer Explanation:**
To solve the problem, we need to determine the marginal revenue at a production level of 60 items.
The marginal revenue \( MR \) is the derivative of the total revenue \( R \) with respect to \( q \). To find the total revenue, we multiply the price function \( p(q) \) by the quantity \( q \):
\[ R(q) = p(q) \cdot q = (-2q + 890) \cdot q \]
\[ R(q) = -2q^2 + 890q \]
Next, we find the derivative of the total revenue function to determine the marginal revenue:
\[ MR(q) = \frac{dR}{dq} = \frac{d}{dq} (-2q^2 + 890q) \]
\[ MR(q) = -4q + 890 \]
Finally, we substitute \( q = 60 \) to find the marginal revenue at this production level:
\[ MR(60) = -4(60) + 890 \]
\[ MR(60) = -240 + 890 \]
\[ MR(60) = 650 \]
Therefore, the marginal revenue at a production level of 60 items is:
\[ \boxed{650} \]
dollars per item.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12d36c83-f991-4c22-88cf-2adf2ca3b459%2F4767c806-0527-4a61-a12f-a2283d2e6cbc%2Fk8ud5xa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Given cost and price (demand) functions \( C(q) = 100q + 43,700 \) and \( p(q) = -2q + 890 \), what is the marginal revenue at a production level of 60 items?
The marginal revenue is \(\_\_\_\_\) dollars per item.
(Round answer to nearest dollar.)
**Answer Explanation:**
To solve the problem, we need to determine the marginal revenue at a production level of 60 items.
The marginal revenue \( MR \) is the derivative of the total revenue \( R \) with respect to \( q \). To find the total revenue, we multiply the price function \( p(q) \) by the quantity \( q \):
\[ R(q) = p(q) \cdot q = (-2q + 890) \cdot q \]
\[ R(q) = -2q^2 + 890q \]
Next, we find the derivative of the total revenue function to determine the marginal revenue:
\[ MR(q) = \frac{dR}{dq} = \frac{d}{dq} (-2q^2 + 890q) \]
\[ MR(q) = -4q + 890 \]
Finally, we substitute \( q = 60 \) to find the marginal revenue at this production level:
\[ MR(60) = -4(60) + 890 \]
\[ MR(60) = -240 + 890 \]
\[ MR(60) = 650 \]
Therefore, the marginal revenue at a production level of 60 items is:
\[ \boxed{650} \]
dollars per item.
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