2) The total cost in dollars of producing x lawn mowers is given by C(x) = 4,000+ 90x- -. Find the x² marginal average cost at x = 20, C'(20) and interpret the result. A) -$1.33; a unit increase in production will decrease the average cost per unit by approximately $1.33 at a production level of 20 units. B) -$20.33; a unit increase in production will decrease the average cost per unit by approximately $20.33 at a production level of 20 units. C) -$10.33; a unit increase in production will decrease the average cost per unit by approximately $10.33 at a production level of 20 units. D) -$13.33; a unit increase in production will decrease the average cost per unit by approximately $13.33 at a production level of 20 units. deled by 2)

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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### Calculus Problem Set

#### 1. Limit Expressions
Evaluate the following limit expressions:

A) \(\frac{(h + 2x)}{x^2(x + h)^2}\)

B) \(\frac{2(h + 2x)}{x^2(x + h)^2}\)

C) \(\frac{-2(h + 2x + xh)}{x^2(x + h)^2}\)

D) \(\frac{2(h + x)}{x^2(x + h)^2}\)

(Correct Answer: B)

#### 2. Marginal Average Cost
The total cost in dollars of producing \(x\) lawn mowers is given by:
\[ C(x) = 4,000 + 90x - \frac{x^2}{3} \]
Find the marginal average cost at \(x = 20\), denoted as \(C'(20)\), and interpret the result.

Options:
- A) \(-$1.33\): A unit increase in production will decrease the average cost per unit by approximately $1.33 at a production level of 20 units.
- B) \(-$20.33\): A unit increase in production will decrease the average cost per unit by approximately $20.33 at a production level of 20 units.
- C) \(-$10.33\): A unit increase in production will decrease the average cost per unit by approximately $10.33 at a production level of 20 units.
- D) \(-$13.33\): A unit increase in production will decrease the average cost per unit by approximately $13.33 at a production level of 20 units.

#### 3. Marginal Revenue
The revenue (in thousands of dollars) from producing \(x\) units of an item is modeled by:
\[ R(x) = 5x - 0.0005x^2 \]
Find the marginal revenue at \(x = 1,000\).

Options:
- A) $10,300.00
- B) $4.00
- C) $4.50
- D) $104.00

(Correct Answer: B)

#### 4. Derivative Calculation
Find the derivative of the expression:
\[ \frac{d}{dx} \left( \frac{4}{x^4} - 5x^{3
Transcribed Image Text:### Calculus Problem Set #### 1. Limit Expressions Evaluate the following limit expressions: A) \(\frac{(h + 2x)}{x^2(x + h)^2}\) B) \(\frac{2(h + 2x)}{x^2(x + h)^2}\) C) \(\frac{-2(h + 2x + xh)}{x^2(x + h)^2}\) D) \(\frac{2(h + x)}{x^2(x + h)^2}\) (Correct Answer: B) #### 2. Marginal Average Cost The total cost in dollars of producing \(x\) lawn mowers is given by: \[ C(x) = 4,000 + 90x - \frac{x^2}{3} \] Find the marginal average cost at \(x = 20\), denoted as \(C'(20)\), and interpret the result. Options: - A) \(-$1.33\): A unit increase in production will decrease the average cost per unit by approximately $1.33 at a production level of 20 units. - B) \(-$20.33\): A unit increase in production will decrease the average cost per unit by approximately $20.33 at a production level of 20 units. - C) \(-$10.33\): A unit increase in production will decrease the average cost per unit by approximately $10.33 at a production level of 20 units. - D) \(-$13.33\): A unit increase in production will decrease the average cost per unit by approximately $13.33 at a production level of 20 units. #### 3. Marginal Revenue The revenue (in thousands of dollars) from producing \(x\) units of an item is modeled by: \[ R(x) = 5x - 0.0005x^2 \] Find the marginal revenue at \(x = 1,000\). Options: - A) $10,300.00 - B) $4.00 - C) $4.50 - D) $104.00 (Correct Answer: B) #### 4. Derivative Calculation Find the derivative of the expression: \[ \frac{d}{dx} \left( \frac{4}{x^4} - 5x^{3
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