The outer perimeter of the window frame is fixed at 6M.

Let the radius be r, and the two sides of the rectangle are 2r, 3-2r respectively.

Area S= 1/2πr? +(3-2r)*2r

Finishing S=( 1/2π-4)r? +6r

Because the quadratic function opens down

When r = 6/(8-π) >: is 0, S gets the maximum value.

Smax= 18/(8-π)