Solution: The perimeter of a rectangle is given by $ P = 2(l + w) $, where $ l $ is length and $ w $ is width. Given $ 2(l + w) = 80 $, we simplify to $ l + w = 40 $. The area $ A = lw $. Express $ w = 40 - l $, so $ A = l(40 - l) = 40l - l^2 $. This quadratic equation opens downward, and its maximum occurs at the vertex $ l = \frac-b2a = \frac-40-2 = 20 $. Substituting $ l = 20 $, $ w = 20 $, so $ A = 20 \times 20 = 400 $. The maximum area is $\boxed400$. - Crosslake