Answer
$$x = 18{\text{in and }}y = 36{\text{in}}$$
Work Step by Step
$$\eqalign{
& {\text{From the given image:}} \cr
& {\text{Let }}V{\text{ be the volume to be maximized}} \cr
& V = {x^2}y{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{We know that the perimeter }}\left( {girth} \right){\text{ is 108inches}},{\text{ then}} \cr
& P = 4x + y \cr
& 4x + y = 108 \cr
& {\text{Solving the previous equation for }}y \cr
& y = 108 - 4x \cr
& {\text{Substitute }}108 - 4x{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& V = {x^2}\left( {108 - 4x} \right) \cr
& V = 108{x^2} - 4{x^3} \cr
& {\text{Differentiate}} \cr
& \frac{{dV}}{{dx}} = 216x - 12{x^2} \cr
& {\text{Let }}\frac{{dV}}{{dx}} = 0,{\text{ then}} \cr
& 216x - 12{x^2} = 0 \cr
& 12x\left( {18 - x} \right) = 0 \cr
& x = 0,{\text{ }}x = 18 \cr
& {\text{Taking }}x = 18 \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}V}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {216x - 12{x^2}} \right] \cr
& \frac{{{d^2}V}}{{d{x^2}}} = 516 - 24x \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}V}}{{d{x^2}}}} \right|_{x = 18}} = 216 - 24\left( {18} \right) < 0{\text{ Relative maximum at }}x = 18 \cr
& {\text{Calculate }}y \cr
& y = 108 - 4x \to y = y = 108 - 4\left( {18} \right) \cr
& y = 36 \cr
& {\text{Therefore, the dimensions are:}} \cr
& x = 18{\text{in and }}y = 36{\text{in}} \cr} $$
