#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.
