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 $2y$ = $\sqrt 2{r}$