Answer
$${\text{length}} = 5\sqrt 2 {\text{ and width}} = \frac{5}{{\sqrt 2 }}$$
Work Step by Step
$$\eqalign{
& {\text{From the given image:}} \cr
& {\text{Let }}A{\text{ be the area to be maximized}} \cr
& A = 2xy{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{We have that }}y = \sqrt {25 - {x^2}} \cr
& {\text{Substituting }}\sqrt {25 - {x^2}} {\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& A = 2x\sqrt {25 - {x^2}} ,{\text{ domain }}\left( {0 < x < 5} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dA}}{{dx}} = 2x\left( {\frac{{ - 2x}}{{2\sqrt {25 - {x^2}} }}} \right) + 2\sqrt {25 - {x^2}} \cr
& \frac{{dA}}{{dx}} = \frac{{ - 2{x^2}}}{{\sqrt {25 - {x^2}} }} + 2\sqrt {25 - {x^2}} \cr
& {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr
& \frac{{ - 2{x^2}}}{{\sqrt {25 - {x^2}} }} + 2\sqrt {25 - {x^2}} = 0 \cr
& \frac{{{x^2}}}{{\sqrt {25 - {x^2}} }} = \sqrt {25 - {x^2}} \cr
& {x^2} = 25 - {x^2} \cr
& 2{x^2} = 25 \cr
& x = \pm \frac{5}{{\sqrt 2 }} \cr
& {\text{Taking }}x = \frac{5}{{\sqrt 2 }} \cr
& {\text{Using the first derivative test}} \cr
& {\text{Interval }}\left( {0,\frac{5}{{\sqrt 2 }}} \right) \cr
& {\left. {\frac{{dA}}{{dx}}} \right|_{x = 3}} = \frac{7}{2} > 0 \cr
& {\text{Interval }}\left( {\frac{5}{{\sqrt 2 }},5} \right) \cr
& {\left. {\frac{{dA}}{{dx}}} \right|_{x = 4}} = - \frac{{14}}{3} < 0 \cr
& {\text{Therefore, there is a relative maximum at }}x = \frac{5}{{\sqrt 2 }} \cr
& y = \sqrt {25 - {x^2}} \to y = \sqrt {25 - {{\left( {\frac{5}{{\sqrt 2 }}} \right)}^2}} = \frac{5}{{\sqrt 2 }} \cr
& {\text{The length}}\,{\text{is }}2x = 2\left( {\frac{5}{{\sqrt 2 }}} \right) = 5\sqrt 2 \cr
& {\text{The width}}\,{\text{is }}y = \frac{5}{{\sqrt 2 }} \cr} $$
