Let the dimensions be L and W. Then P = perimeter = 800 m = 2W + L.
Constraint:
We want to maximize the area. This area is L*W. We can eliminate one variable or the other by solving 800=2W + L for either. L = 800 - 2W. Then
the area is A(W) = (800 - 2W)*W.
Differentiate this with respect to W: A '(W) = (800 - 2W)(1) + W(-2), or
800 - 2W - 2W = 800 - 4W
Set this derivative = to 0 and solve for W: W = 800/4, or W = 200. Then L = 800 - 2W becomes L = 800 - 2(200) = 400.
Thus, W = 200 m and L = 400. From this data we get max area = (200)(400) m^2, or 80000 m^2.