Answer
$${\text{Width}} = {\text{length}} = \sqrt A {\text{cm}}$$
Work Step by Step
$$\eqalign{
& {\text{From the image shown below we know that the perimeter is}} \cr
& {\text{given by the equation}} \cr
& xy = A{\text{ }}\left( {\bf{1}} \right) \cr
& P = 2x + 2y{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Solve equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& y = \frac{A}{x} \cr
& {\text{Substitute }}\frac{A}{x}{\text{ into equation }}\left( {\bf{2}} \right) \cr
& P = 2x + 2\left( {\frac{A}{x}} \right) \cr
& P = 2x + \frac{{2A}}{x} \cr
& {\text{Differentiate}} \cr
& \frac{{dP}}{{dx}} = 2 - \frac{{2A}}{{{x^2}}} \cr
& {\text{Find the critical points by solving }}\frac{{dP}}{{dx}} = 0 \cr
& 2 - \frac{{2A}}{{{x^2}}} = 0 \cr
& {x^2} = A \cr
& x = \pm \sqrt A \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}P}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {2 - \frac{{2A}}{{{x^2}}}} \right] \cr
& \frac{{{d^2}P}}{{d{x^2}}} = \frac{{4A}}{{{x^3}}} \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}P}}{{d{x^2}}}} \right|_{x = - \sqrt A }} = \frac{{4A}}{{{{\left( { - \sqrt A } \right)}^3}}} < 0{\text{ Relative maximum}} \cr
& {\left. {\frac{{{d^2}P}}{{d{x^2}}}} \right|_{x = \sqrt A }} = \frac{{4A}}{{{{\left( {\sqrt A } \right)}^3}}} > 0{\text{ Relative minimum}} \cr
& {\text{Taking }}x = \sqrt A \cr
& y = \frac{A}{x} \to y = \frac{A}{{\sqrt A }} = \sqrt A \cr
& {\text{Therefore, the numbers are: }}x = \sqrt A {\text{ and }}y = \sqrt A \cr
& {\text{The dimensions are Width}} = {\text{length}} = \sqrt A {\text{cm}} \cr} $$
