Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.7 Optimization Problems - 3.7 Exercises - Page 265: 25

Answer

$\sqrt 2{r}$ and $\sqrt 2{r}$

Work Step by Step

Let's note by $2x$ and $2y$ the dimensions of the rectangle (see figure). The area of the rectangle is $A=(2x)(2y)$ = $4xy$. We also have, using the Pythagorean Theorem: $x^{2}+y^{2}$ = $r^{2}$ $y$ = $\sqrt {r^{2}-x^{2}}$ so that area is $A(x)$ = $4x\sqrt {r^{2}-x^{2}}$ $A'(x)$ = $4\left(\sqrt {r^{2}-x^{2}}-\frac{x^{2}}{\sqrt {r^{2}-x^{2}}}\right)$ = $\frac{4(r^{2}-2x^{2})}{\sqrt {r^{2}-x^{2}}}$ The critical number is $x$ = $\frac{r}{\sqrt 2}$ Clearly this gives a maximum $y$ = $\sqrt {r^{2}-\left(\frac{r}{\sqrt 2}\right)^{2}}$ = $\frac{r}{\sqrt 2}$ = $x$ $x$ = $y$ so the rectangle is a square. The dimensions are $2x$ = $\sqrt 2{r}$ and $2y$ = $\sqrt 2{r}$
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