Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.4 - The Derivative as a Rate of Change - Exercises 3.4 - Page 136: 23

Answer

a. $110$ dollars. b. $80$ dollars. c. $79.9\approx80$ dollars.

Work Step by Step

a. Given the cost equation $c(x)=2000+100x-0.1x^2$, we can obtain the cost for producing the first 100 washing machines is $c(100)=2000+100(100)-0.1(100)^2=11000$ Thus the average cost per machine is $\bar c=11000/100=110$ dollars. b. The marginal cost when 100 washing machines are produced is the value of the derivative of the cost function at that point. We have $c'(x)=100-0.2x$ and $c'(100)=100-0.2(100)=80$ dollars. c. Let $x=101$; we have $c(101)=2000+100(101-0.1(101)^2=11079.9$ dollars. Thus the cost of producing one more washing machine after the first 100 have been made is: $c(101)-c(100)=11079.9-11000=79.9\approx80$ dollars and this is the marginal cost.
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