Answer
$$x = 6\sqrt 2 {\text{ and }}y = 2\sqrt 2 $$
Work Step by Step
$$\eqalign{
& {\text{From the image shown below:}} \cr
& {\text{Let }}A{\text{ be the area to be maximized}} \cr
& A = \left( {2x} \right)\left( {2y} \right){\text{ }} \cr
& A = 4xy{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{We know that the equation of the ellipse is }}\frac{{{x^2}}}{{144}} + \frac{{{y^2}}}{{16}} = 1 \cr
& {\text{Solving for }}y \cr
& {x^2} + 9{y^2} = 144 \cr
& {y^2} = \frac{{144 - {x^2}}}{9} \cr
& y = \pm \frac{{\sqrt {144 - {x^2}} }}{3} \cr
& {\text{Taking the values in the first quadrant}} \cr
& y = \frac{{\sqrt {144 - {x^2}} }}{3},{\text{ Domain }}0 < x < 12 \cr
& {\text{Substitute }}\frac{{\sqrt {144 - {x^2}} }}{3}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& A = 4x\left( {\frac{{\sqrt {144 - {x^2}} }}{3}} \right) \cr
& A = \frac{4}{3}x\sqrt {144 - {x^2}} \cr
& {\text{Differentiate using the product rule}} \cr
& \frac{{dA}}{{dx}} = \frac{4}{3}x\left( {\frac{{ - 2x}}{{2\sqrt {144 - {x^2}} }}} \right) + \frac{4}{3}\sqrt {144 - {x^2}} \cr
& \frac{{dA}}{{dx}} = - \frac{{4{x^2}}}{{3\sqrt {144 - {x^2}} }} + \frac{4}{3}\sqrt {144 - {x^2}} \cr
& {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr
& - \frac{{4{x^2}}}{{3\sqrt {144 - {x^2}} }} + \frac{4}{3}\sqrt {144 - {x^2}} = 0 \cr
& \sqrt {144 - {x^2}} = \frac{{{x^2}}}{{\sqrt {144 - {x^2}} }} \cr
& 144 - {x^2} = {x^2} \cr
& 2{x^2} = 144 \cr
& {x^2} = 72 \cr
& x = \pm 6\sqrt 2 {\text{, }}6\sqrt 2 {\text{ is in the domain, then}} \cr
& x = 6\sqrt 2 \cr
& {\text{Using the first derivative test}} \cr
& f'\left( 8 \right) = \frac{{16\sqrt 5 }}{{15}} > 0,{\text{ }}f'\left( 9 \right) = - \frac{{8\sqrt 7 }}{7} < 0 \cr
& {\text{The derivative changes + to }} - {\text{ at }}x = 6\sqrt 2 ,{\text{ then}} \cr
& {\text{There is a maximum relative at }}x = 6\sqrt 2 \cr
& {\text{Calculate }}y \cr
& y = \frac{{\sqrt {144 - {x^2}} }}{3} \to y = \frac{{\sqrt {144 - {{\left( {6\sqrt 2 } \right)}^2}} }}{3} = 2\sqrt 2 \cr
& {\text{Therefore, the dimensions are:}} \cr
& x = 6\sqrt 2 {\text{ and }}y = 2\sqrt 2 \cr} $$
