Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 217: 41

Answer

a. $\frac{8}{35}$ b. $\frac{31}{240}$ c. $\frac{3x+2}{x^2+4x}$ d. $\frac{-3x^2-4x-8}{(x^2+4x)^2}$

Work Step by Step

We are given $C(x)=\frac{3x+2}{x+4}$ The average cost is defined by: $\bar{C}(x)=\frac{C(x)}{x}=\frac{3x+2}{x+4}.\frac{1}{x}=\frac{3x+2}{x^2+4x}$ a. 10 units then the average cost is: $\bar{C}(x)=\frac{3x+2}{x^2+4x}=\frac{3(10)+2}{10^2+4(10)}=\frac{8}{35}$ b. 20 units then the average cost is: $\bar{C}(x)=\frac{3x+2}{x^2+4x}=\frac{3(20)+2}{20^2+4(20)}=\frac{31}{240}$ c. x units then the average cost is: $\bar{C}(x)=\frac{3x+2}{x^2+4x}$ d. The marginal average cost is given by: $\frac{d}{dx}[\bar{C}(x)]=\frac{3(x^2+4x)-(2x+4)(3x+2)}{(x^2+4x)^2}$ $=\frac{3x^2+12x-6x^2-4x-12x-8}{(x^2+4x)^2}$ $=\frac{-3x^2-4x-8}{(x^2+4x)^2}$
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