Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises: 14

Answer

The box should be $40\times 40 \times 20$ to minimize the volume.

Work Step by Step

Let $b$ be the length of the base and $h$ the height. The volume is $32,000=b^{2}h$, so $h=\frac{32,000}{b^{2}}$. The surface area of the open box is $S=b^{2}+4hb=b^{2}+4\frac{32,000}{b}$. $S'(b)=2b-\frac{128,000}{b^{2}}$, so $S'(b)$ is $0$ when $b=40$. This gives an absolute minimum, so the two bases should be $40$ and the height should be $20$.
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