The cost, in thousands of dollars, of airing x television commercials during a sports event is given by C(x) = 150 + 2,600x – 0.06x². (a) Find the marginal cost function C'(x). C'(x) = (b) Use the marginal cost to approximate the cost to air the 5th commercial. Convert your answer to dollars. The cost to air the 5th commercial is approximately × dollars. (c) What is the exact cost to air the 5th commercial? Convert your answer to dollars. The exact cost to air the 5th commercial is × dollars.

ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN:9780190931919
Author:NEWNAN
Publisher:NEWNAN
Chapter1: Making Economics Decisions
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### Cost Function and Marginal Analysis

#### Problem Statement

The cost, in thousands of dollars, of airing \( x \) television commercials during a sports event is given by:

\[ C(x) = 150 + 2.600x - 0.06x^2. \]

#### Questions

**(a) Find the marginal cost function \( C'(x) \).**

\[ C'(x) = \text{(Enter marginal cost function here)} \]

**(b) Use the marginal cost to approximate the cost to air the 5th commercial. Convert your answer to dollars.**

The cost to air the 5th commercial is approximately \(\boxed{\text{(enter approximate cost here)}}\) dollars.

**(c) What is the exact cost to air the 5th commercial? Convert your answer to dollars.**

The exact cost to air the 5th commercial is \(\boxed{\text{(enter exact cost here)}}\) dollars.

#### Explanation:

- **Part (a):** Calculate the marginal cost function, which is the derivative of the cost function. This represents the cost change with respect to the number of commercials aired.

- **Part (b):** Use the marginal cost function to estimate the cost of the additional (5th) commercial. This approximation is common in economic analysis to simplify calculations.

- **Part (c):** Compute the exact cost of airing the 5th commercial by evaluating the difference in cost between airing 5 and 4 commercials.

This exercise involves differentiation and basic cost analysis, core concepts in calculus and economics.
Transcribed Image Text:### Cost Function and Marginal Analysis #### Problem Statement The cost, in thousands of dollars, of airing \( x \) television commercials during a sports event is given by: \[ C(x) = 150 + 2.600x - 0.06x^2. \] #### Questions **(a) Find the marginal cost function \( C'(x) \).** \[ C'(x) = \text{(Enter marginal cost function here)} \] **(b) Use the marginal cost to approximate the cost to air the 5th commercial. Convert your answer to dollars.** The cost to air the 5th commercial is approximately \(\boxed{\text{(enter approximate cost here)}}\) dollars. **(c) What is the exact cost to air the 5th commercial? Convert your answer to dollars.** The exact cost to air the 5th commercial is \(\boxed{\text{(enter exact cost here)}}\) dollars. #### Explanation: - **Part (a):** Calculate the marginal cost function, which is the derivative of the cost function. This represents the cost change with respect to the number of commercials aired. - **Part (b):** Use the marginal cost function to estimate the cost of the additional (5th) commercial. This approximation is common in economic analysis to simplify calculations. - **Part (c):** Compute the exact cost of airing the 5th commercial by evaluating the difference in cost between airing 5 and 4 commercials. This exercise involves differentiation and basic cost analysis, core concepts in calculus and economics.
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