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Given Cost and Revenue functions C(q) =q³-9q² +53q + 5000 and R(q) = -3q² + 2500q, what is the marginal profit at a production level of 60 items? The marginal profit is dollars per item.

Given Cost and Revenue functions C(q) =q³-9q² +53q + 5000 and R(q) = -3q² + 2500q, what is the marginal profit at a production level of 60 items? The marginal profit is dollars per item.

Calculus: Early Transcendentals
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 Statement Given the Cost and Revenue functions: \[ C(q) = q^3 - 9q^2 + 53q + 5000 \] \[ R(q) = -3q^2 + 2500q \] what is the marginal profit at a production level of 60 items? ### Explanation To find the marginal profit at a production level of 60 items, we need to differentiate the Cost and Revenue functions with respect to \( q \) and then evaluate the difference at \( q = 60 \). ### Solution Steps 1. **Differentiate the Cost Function**: \[ C'(q) = \frac{d}{dq} (q^3 - 9q^2 + 53q + 5000) = 3q^2 - 18q + 53 \] 2. **Differentiate the Revenue Function**: \[ R'(q) = \frac{d}{dq} (-3q^2 + 2500q) = -6q + 2500 \] 3. **Find the Marginal Profit Function**: \[ \text{Marginal Profit} = R'(q) - C'(q) \] 4. **Evaluate at \( q = 60 \)**: \[ \text{Marginal Profit at } q = 60 \] \[ = ( -6(60) + 2500 ) - ( 3(60)^2 - 18(60) + 53 ) \] \[ = (-360 + 2500) - (10800 - 1080 + 53) \] \[ = 2140 - 9773 \] \[ = -7633 \] ### Answer The marginal profit is -7633 dollars per item at a production level of 60 items.
Transcribed Image Text:### Problem Statement Given the Cost and Revenue functions: \[ C(q) = q^3 - 9q^2 + 53q + 5000 \] \[ R(q) = -3q^2 + 2500q \] what is the marginal profit at a production level of 60 items? ### Explanation To find the marginal profit at a production level of 60 items, we need to differentiate the Cost and Revenue functions with respect to \( q \) and then evaluate the difference at \( q = 60 \). ### Solution Steps 1. **Differentiate the Cost Function**: \[ C'(q) = \frac{d}{dq} (q^3 - 9q^2 + 53q + 5000) = 3q^2 - 18q + 53 \] 2. **Differentiate the Revenue Function**: \[ R'(q) = \frac{d}{dq} (-3q^2 + 2500q) = -6q + 2500 \] 3. **Find the Marginal Profit Function**: \[ \text{Marginal Profit} = R'(q) - C'(q) \] 4. **Evaluate at \( q = 60 \)**: \[ \text{Marginal Profit at } q = 60 \] \[ = ( -6(60) + 2500 ) - ( 3(60)^2 - 18(60) + 53 ) \] \[ = (-360 + 2500) - (10800 - 1080 + 53) \] \[ = 2140 - 9773 \] \[ = -7633 \] ### Answer The marginal profit is -7633 dollars per item at a production level of 60 items.
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