Answer
$${\text{Width}} = {\text{length}} = \frac{1}{4}P{\text{ units}}$$
Work Step by Step
$$\eqalign{
& {\text{From the image shown below we know that the perimeter is}} \cr
& {\text{given by the equation}} \cr
& 2x + 2y = P{\text{ }}\left( {\bf{1}} \right) \cr
& A = xy{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Solve equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& 2y = P - 2x \cr
& {\text{ }}y = \frac{1}{2}P - x \cr
& {\text{Substitute }}\frac{1}{2}P - x{\text{ into equation }}\left( {\bf{2}} \right) \cr
& A = x\left( {\frac{1}{2}P - x} \right) \cr
& \frac{{dA}}{{dx}} = \frac{1}{2}Px - {x^2} \cr
& {\text{Differentiate}} \cr
& \frac{{dA}}{{dx}} = \frac{1}{2}P - 2x \cr
& {\text{Find the critical points by solving }}\frac{{dA}}{{dx}} = 0 \cr
& \frac{1}{2}P - 2x = 0 \cr
& x = \frac{1}{4}P \cr
& {\text{Calculate the second derivative}} \cr
& \frac{{{d^2}A}}{{d{x^2}}} = - 2 \cr
& {\text{By the second derivative test:}} \cr
& {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = 20}} = - 2 < 0{\text{ Relative maximum}} \cr
& {\text{Taking }}x = \frac{1}{4}P \cr
& {\text{ }}y = \frac{1}{2}P - x \to y = \frac{1}{2}P - \frac{1}{4}P = \frac{1}{4}P \cr
& {\text{Therefore, the numbers are: }}x = \frac{1}{4}P{\text{ and }}y = \frac{1}{4}P \cr
& {\text{The dimensions are Width}} = {\text{length}} = \frac{1}{4}P{\text{ units}} \cr} $$
