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 - Page 337: 8

Answer

The dimensions of the rectangle with minimum perimeter are $x=y=10\sqrt 10$ m.

Work Step by Step

If the rectangle has dimensions $x$ and $y$, then its area is $xy=1000 m^{2}$, so $y=\frac{1000}{x}$. The perimeter function is $y=2x+2y = 2x+\frac{2000}{x}$, and we wish to minimize this. $P'(x)=2-\frac{2000}{x^{2}}$, so the only critical number is $x=10\sqrt 10$. To solve for $y$, we use $y=\frac{1000}{x}$, and find $y=10\sqrt 10$ as well, giving us the dimensions that minimize the perimeter.
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