Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 12 - Section 12.2 - Applications of Maxima and Minima - Exercises - Page 888: 39

Answer

$4$

Work Step by Step

We are given that $V=250 \ cm^3$ Volume can be computed as: $V=\pi r^2 h$ or, $h=\dfrac{250}{\pi r^2}$ Next, the surface area is: $S=2\pi r^2+2\pi rh$ Cost $=2 \pi r^2 (0.02)+2\pi r \times \dfrac{250}{\pi r^2}\times 0.01=0.04 \pi r^2 +\dfrac{5}{r}$ For minimum cost, we must have $\dfrac{dC}{dr}=0$ or, $0.08 \pi r -\dfrac{5}{r^2}=0$ or, $r=2.70 \ cm$ Thus, $h=\dfrac{250}{\pi (2.70)^2}=10.92 \ cm$ Now, $\dfrac{h}{r}=\dfrac{10.92}{2.7}=4$
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