Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.5 - Applied Optimization - Exercises 4.5 - Page 221: 2

Answer

Local maximum value at $x=2$ for a square with width and length is 2 in.

Work Step by Step

Let $x$ and $y$ be the sides of a rectangle with $xy=16$ and perimeter $2x+2y=8$ Thus, we have $f(x)=x(4-x)$ and $f'(x)=4-2x$ Need to find the maximal value of $f(x)$ . when $f'(x)=0$ then $x=2$ This implies that the local maximum is at $x=2$ when $x \gt 0$. $x=2$ and $y=4-2=2$ , that is, for a square and perimeter $= (4)(4)=16 in$ Hence, Local maximum value at $x=2$ for a square with width and length is 2 in.
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