#L = l_s+l_e# where

#l_s# is the piece associated to the square.

#l_e# is the piece associated to the equilateral triangle.

The associated areas are:

#A_s = (l_s/4)^2 = l_s^2/16# is the square area

#A_e = 1/2(l_e/3)(l_e/3)sqrt(3)/2 = l_e^2sqrt(3)/36#

so the total area is

#A = A_s+A_e#

so substituting for #l_e = L - l_s#

#A = l_s^2/16+(L-l_s)^2sqrt(3)/36#

The maximum is attained at #l_s=l_s^0# such that

#(dA)/(dl_s)]_(l_s=l_s^0) = 2l_s^0/16-2(L-l_s^0)sqrt(3)/36 = 0#

Solving for #l_s# we get

#l_s = (4 sqrt[3] L)/(9 + 4 sqrt[3]) approx 4.34965#

and

#l_e=10-4.34965=5.65035#

This solution is a minimum solution because

#(d^2A)/(dl_s^2)=1/8 + 1/(6 sqrt[3]) > 0# characterizing a minimum.