jmc

Since the perimeter is 40, the sides of the rectangle add up to $40/2=20.$ Let $x$ be one side length of the rectangle. Then the other side length is $20-x,$ so the area is $$x(20-x)=20x-x^2.$$Completing the square, we get $$-x^2+20x=-x^2+20x-100+100=100-(x-10{)}^2.$$Thus, the maximum area of the rectangle is $\boxed{100}$ square feet, which occurs for a $10\times 10$ square.