# Question 1b8d8

Apr 17, 2017

See below.

#### Explanation:

$L = {l}_{s} + {l}_{e}$ where

${l}_{s}$ is the piece associated to the square.
${l}_{e}$ is the piece associated to the equilateral triangle.

The associated areas are:

${A}_{s} = {\left({l}_{s} / 4\right)}^{2} = {l}_{s}^{2} / 16$ is the square area

${A}_{e} = \frac{1}{2} \left({l}_{e} / 3\right) \left({l}_{e} / 3\right) \frac{\sqrt{3}}{2} = {l}_{e}^{2} \frac{\sqrt{3}}{36}$

so the total area is

$A = {A}_{s} + {A}_{e}$

so substituting for ${l}_{e} = L - {l}_{s}$

$A = {l}_{s}^{2} / 16 + {\left(L - {l}_{s}\right)}^{2} \frac{\sqrt{3}}{36}$

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

${l}_{s} = \frac{4 \sqrt{3} L}{9 + 4 \sqrt{3}} \approx 4.34965$

and

${l}_{e} = 10 - 4.34965 = 5.65035$

This solution is a minimum solution because

$\frac{{d}^{2} A}{{\mathrm{dl}}_{s}^{2}} = \frac{1}{8} + \frac{1}{6 \sqrt{3}} > 0$ characterizing a minimum.