Step 1 :Let the length of the plot be \(l\) and the width be \(w\). We have the equation \(l+2w=200\) because only three sides need to be fenced.
Step 2 :Rearrange the equation to find \(l=200-2w\).
Step 3 :We want to maximize the area of this rectangular plot, which is given by \(lw\). Substituting \(l\) into our expression for area, we have \[(200-2w)(w)=200w-2w^2\].
Step 4 :We will now complete the square to find the maximum value of this expression. Factoring a \(-2\) out, we have \[-2(w^2-100w)\].
Step 5 :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 \[-2(w^2-100w+2500-2500) \Rightarrow -2(w-50)^2+5000\].
Step 6 :Since the maximum value of \(-(w-50)^2\) is 0 (perfect squares are always nonnegative), the maximum value of the entire expression is 5000, which is achieved when \(w=50\) and \(l=200-2w=100\).
Step 7 :Thus, the maximum area of the plot is \(\boxed{5000}\) square feet.