University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 146: 23

Answer

a) The average cost per machine is $110$ dollars. b) The marginal cost is $80$ dollars. c) See down for calculations and proof.

Work Step by Step

$$c(x)=2000+100x-0.1x^2$$ a) To find the average cost per machine, we divide the cost $c(x)$ by the number of machines $x$. Here, the average cost per machine of producing the first $100$ machines is $$\frac{c(100)}{100}=\frac{2000+100\times100-0.1\times100^2}{100}=\frac{11000}{100}=110(dollars)$$ b) The marginal cost is the derivative of the cost of producing $x$ washing machines $c(x)$. So we would call the marginal cost here $c'(x)$. $$c'(x)=100-0.2x$$ The marginal cost when $100$ washing machines are produced is $$c'(100)=100-0.2\times100=80(dollars)$$ c) The cost of producing $101$ washing machines is $$c(101)=2000+100\times101-0.1\times101^2=11079.9(dollars)$$ The cost of producing $100$ washing machines is $$c(100)=2000+100\times100-0.1\times100^2=11000(dollars)$$ So, the cost of producing one more machine after $100$ machines have been made is $$c(101)-c(100)=11079.9-11000=79.9(dollars)$$ It is quite the same as the marginal cost when $100$ washing machines are produced.
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