jmc

Let the length of the playground be $l$ and the width be $w$. We have the equation $2l+2w=200\Rightarrow l+w=100$. We want to maximize the area of this rectangular playground, which is given by $lw$. From our equation, we know that $l=100-w$. Substituting this in to our expression for area, we have $$(100-w)(w)=100w-w^2$$We will now complete the square to find the maximum value of this expression. Factoring a $-1$ out, we have $$-(w^2-100w)$$In order for the expression inside the parenthesis to be a perfect square, we need to add and subtract $(100/2{)}^2=2500$ inside the parenthesis. Doing this, we get $$-(w^2-100w+2500-2500)\Rightarrow -(w-50{)}^2+2500$$Since the maximum value of $-(w-50{)}^2$ is 0 (perfect squares are always nonnegative), the maximum value of the entire expression is 2500, which is achieved when $w=50$ and $l=100-w=50$ (the playground is a square). Thus, the maximum area of the playground is $\boxed{2500}$ square feet.