Let the upper base #y# of the rectangle be the segment of a line parallel to the base of the equilateral triangle at an unknown distance x from it. In such a way the triangle is divided in two triangles, the equilateral one having height #h=Lroot2{3}/2# and a smaller one having height #h_1=Lroot2{3}/2-x#, that are similar! so we can write the proportion #\frac{L}{y}=\frac{Lroot2{3}/2}{Lroot2{3}/2-x}#. By insulating the #y# we obtain #y=L-\frac{2}{root2{3}}x#
The rectangle area is #S(x,y)=x*y# but #S(x)=x*(L-\frac{2}{root2{3}}x)=Lx-\frac{2}{root2{3}}x^2#.
By deriving #S(x)# we get #S'(x)=L-\frac{4}{root2{3}}x# whose root is #x=L\frac{root2{3}}{4}# and consequently #y=L-\frac{2}{root2{3}}\frac{root2{3}}{4}L=\frac{L}{2}#
