Answer
$${\text{length}} = \sqrt 2 r{\text{ and width}} = \frac{r}{{\sqrt 2 }}$$
Work Step by Step
$$\eqalign{
& {\text{From the given image:}} \cr
& {\text{Let }}A{\text{be the area to be maximized}} \cr
& A = 2xy{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{We have that }}y = \sqrt {{r^2} - {x^2}} \cr
& {\text{Substituting }}\sqrt {{r^2} - {x^2}} {\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& A = 2x\sqrt {{r^2} - {x^2}} ,{\text{ domain }}\left( {0 < x < r} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dA}}{{dx}} = 2x\left( {\frac{{ - 2x}}{{2\sqrt {{r^2} - {x^2}} }}} \right) + 2\sqrt {{r^2} - {x^2}} \cr
& \frac{{dA}}{{dx}} = \frac{{ - 2{x^2}}}{{\sqrt {{r^2} - {x^2}} }} + 2\sqrt {{r^2} - {x^2}} \cr
& {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr
& \frac{{ - 2{x^2}}}{{\sqrt {{r^2} - {x^2}} }} + 2\sqrt {{r^2} - {x^2}} = 0 \cr
& \frac{{{x^2}}}{{\sqrt {{r^2} - {x^2}} }} = \sqrt {{r^2} - {x^2}} \cr
& {x^2} = {r^2} - {x^2} \cr
& 2{x^2} = {r^2} \cr
& x = \pm \frac{r}{{\sqrt 2 }} \cr
& {\text{Taking }}x = \frac{r}{{\sqrt 2 }} \cr
& {\text{Using the first derivative test}} \cr
& {\text{Interval }}\left( {0,\frac{r}{{\sqrt 2 }}} \right),{\text{ }}\left. {\frac{{dA}}{{dx}}} \right| > 0 \cr
& {\text{Interval }}\left( {\frac{5}{{\sqrt 2 }},5} \right),{\text{ }}\left. {\frac{{dA}}{{dx}}} \right| < 0 \cr
& {\text{Therefore, there is a relative maximum at }}x = \frac{r}{{\sqrt 2 }} \cr
& y = \sqrt {{r^2} - {x^2}} \to y = \sqrt {{r^2} - {{\left( {\frac{r}{{\sqrt 2 }}} \right)}^2}} = \frac{r}{{\sqrt 2 }} \cr
& {\text{The length}}\,{\text{is }}2x = 2\left( {\frac{r}{{\sqrt 2 }}} \right) = \frac{{2r}}{{\sqrt 2 }} = \sqrt 2 r \cr
& {\text{The width}}\,{\text{is }}y = \frac{r}{{\sqrt 2 }} \cr} $$
