Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 51

Answer

(a) The average rate of change of C with respect to x (i) from $x=100$ to $x=105$: $20.25do/un$ (ii) from $x=100$ to $x=101$: $20.05do/un$ (b) The instantaneous rate of change of C with respect to x when $x=100$ is $20do/un$.

Work Step by Step

$$C(x)=5000+10x+0.05x^2$$ (a) For $x=100$, $C(100)=5000+10\times100+0.05\times100^2=6500(do)$ For $x=101$, $C(101)=5000+10\times101+0.05\times101^2=6520.05(do)$ For $x=105$, $C(105)=5000+10\times105+0.05\times105^2=6601.25(do)$ Now we find the average rate of change of C with respect to x when production level is changed (i) from $x=100$ to $x=105$ The average rate of change is $$\frac{C(105)-C(100)}{105-100}=\frac{6601.25-6500}{5}=20.25(do/un)$$ (i) from $x=100$ to $x=101$ The average rate of change is $$\frac{C(101)-C(100)}{101-100}=\frac{6520.05-6500}{1}=20.05(do/un)$$ The unit is $do/un$ (dollars/unit) (b) Find the derivative of $C(x)$ at $x=a$: $$C'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $$C'(a)=\lim\limits_{x\to a}\frac{(5000+10x+0.05x^2)-(5000+10a+0.05a^2)}{x-a}$$ $$C'(a)=\lim\limits_{x\to a}\frac{10(x-a)+0.05(x^2-a^2)}{x-a}$$ $$C'(a)=\lim\limits_{x\to a}\frac{10(x-a)+0.05(x-a)(x+a)}{x-a}$$ $$C'(a)=\lim\limits_{x\to a}\frac{(x-a)[10+0.05(x+a)]}{x-a}$$ $$C'(a)=\lim\limits_{x\to a}[10+0.05(x+a)]$$ $$C'(a)=10+0.05(a+a)$$ $$C'(a)=10+0.1a$$ The instantaneous rate of change of C with respect to x when $x=100$ is in fact the derivative of C at $x=100$, which is $$C'(100)=10+0.1\times100=20$$ In conclusion, the instantaneous rate of change of C with respect to x when $x=100$ is $20do/un$.
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