Calculus with Applications (10th Edition)

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

Chapter 8 - Further Techniques and Applications of Integration - 8.2 Volume and Average Value - 8.2 Exercises - Page 440: 37

Answer

$$ N(t)=600-20\sqrt {30t} $$ The average value of a function $N(t)$ on the interval$[0,30]$ is given by: $$ \begin{aligned} \text{The average value}&=\frac{1}{b-a} \int_{a}^{b}(N(t)) d t\\ &=\frac{1}{30-0} \int_{0}^{30}(600-20\sqrt {30t} )d t\\ &=200 \mathrm{cases}\\ \end{aligned} $$ Therefore the average daily inventory is $200 \mathrm{cases}$.

Work Step by Step

$$ N(t)=600-20\sqrt {30t} $$ The average value of a function $N(t)$ on the interval$[0,30]$ is given by: $$ \begin{aligned} \text{The average value}&=\frac{1}{b-a} \int_{a}^{b}(N(t)) d t\\ &=\frac{1}{30-0} \int_{0}^{30}(600-20\sqrt {30t} )d t\\ &=\left.\frac{1}{30}\left(600 t-20 \sqrt{30} \cdot \frac{2}{3} t^{3 / 2}\right)\right|_{0} ^{30} \\ &=\left.\frac{1}{30}\left(600 t-\frac{40 \sqrt{30}}{3} t^{3 / 2}\right)\right|_{0} ^{30} \\ &=\frac{1}{30}(18,000-12,000)\\ &=200 \mathrm{cases}\\ \end{aligned} $$ Therefore the average daily inventory is $200 \mathrm{cases}$.
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