Here's what I came up with:
We want to maximize shaded area R, which is the difference in area between the large rectangle and the two squares. R = |DC|*|CB| = sqrt(2)*|EC|*|CB|
The perimeter of triangle BCE is greatest when |EC| = |CB|, and |EB| is constant, so this maximizes |EC| + |CB|. By AM/GM, |EC|*|CB| and R also greatest when the triangle is isosceles. The area of the larger square is 1/2 + sqrt(20)/20.