problem: Richard is building a rectangular playground from 200 feet of fencing. The fencing must entirely enclose the playground. What is the maximum area of this playground?
solution: 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.