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: 24


a) The marginal revenue is $2$ dollars. b) The increase in revenue is $1.96$ dollars. c) $\lim_{x\to\infty}r'(x)=0$

Work Step by Step

$$r(x)=20000\Big(1-\frac{1}{x}\Big)=20000-\frac{20000}{x}$$ a) The marginal revenue is the derivative of the revenue from selling $x$ washing machines $r(x)$. So we would call the marginal revenue here $r'(x)$. $$r'(x)=0-\Big(-\frac{20000}{x^2}\Big)=\frac{20000}{x^2}$$ The marginal revenue when $100$ washing machines are produced is $$r'(100)=\frac{20000}{100^2}=2(dollars)$$ b) The marginal revenue when $101$ washing machines are produced is $$r'(101)=\frac{20000}{101^2}\approx1.961(dollars)$$ This number, or the marginal revenue for $101$ machines, represents the change in the revenue as we increase the production from $100$ machines to $101$ machines. In other words, increasing production to $101$ machines a week will increase the revenue by $1.961$ dollars. c) $$\lim_{x\to\infty}r'(x)=\lim_{x\to\infty}\frac{20000}{x^2}=0$$ As $x\to\infty$, $20000/x^2$ will approach $0$. This shows us that even if we have the capacity to increase production infinitely, the revenue will not increase continuously. There will come a point when it we will stop gaining revenue anymore.
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