Inspired by "Squeeze it in"

Since a triangle BEF has three vertices and a square ABCD four sides, the maximum triangle must have one vertex touching two sides of the square. This is only possible if one vertex of the triangle is also the vertex of the square . Let this be B.
The vertex angles of an equilateral triangles are 60 degrees, so at B 30 degrees are left to be shared by the pair of a triangle sides and a square sides. However sides of an equilateral triangle are equal this is only possible if 30 degrees are shared equally.
$So \angle~ ABF = \angle~ EBC= 15. ~~ \Delta s ~ ABF\equiv EBC, ~ \implies~\Delta~ BEF~is ~isosceles~ with\\ \angle~EBF ~= ~60. ~\\ So ~it ~is ~ equilateral~ with~ side~ BE=\dfrac {BC}{Cos15}=\dfrac {30}{\frac{\sqrt6+\sqrt2} 4}= 30(\sqrt6 - \sqrt2).\\ \therefore ~x+y=6+2=8$

What I boxed in the picture can be rewritten as $30 \sqrt{6} - 30 \sqrt{2}$ , and $6+2 = \boxed{8}$ .