Triangular Prisms.

$\triangle{AEC}$ is an isosceles triangle $\implies EM$ is the perpendicular bisector of base $AC$ $\implies \triangle{MEC}$ is a right triangle and $AM = MC$ .

Since $\angle{ACB}$ is a common angle to both right triangle $ABC$ and $MEC \implies \triangle{ABC} \sim \triangle{MEC} \implies \dfrac{2m}{m} = \dfrac{x + a}{a} \implies x = a$ and the pythagorean theorem in $\triangle{ABC} \implies y = 2m = \sqrt{3}a$ .

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The surface area $S = \sqrt{3}a^2 + (3 + \sqrt{3})ah$ and the volume $V = \dfrac{\sqrt{3}}{2}a^2h = k \implies h = \dfrac{2k}{\sqrt{3}a^2} \implies S(a) = \sqrt{3}a^2 + \dfrac{2(3 + \sqrt{3})k}{\sqrt{3}a} \implies$

$\dfrac{dS}{da} = \dfrac{6a^3 - 2(3 + \sqrt{3})k}{\sqrt{3}a^2} = 0 \:\ a \neq 0 \implies$ $a = \displaystyle (\dfrac{\sqrt{3} + 1}{\sqrt{3}}k)^{\frac{1}{3}} \implies$ $h = \displaystyle (\dfrac{4k}{\sqrt{3}(2 + \sqrt{3})})^{\frac{1}{3}}$

$\implies S = \displaystyle 3\sqrt{3}(\dfrac{\sqrt{3} + 1}{\sqrt{3}})^{\frac{2}{3}}\displaystyle k^{\frac{2}{3}} = \displaystyle k^{\frac{4}{3}}$

Let $j = 3\sqrt{3}(\dfrac{\sqrt{3} + 1}{\sqrt{3}})^{\frac{2}{3}}$

$\implies \displaystyle k^{\frac{2}{3}}(j - k^{\frac{2}{3}}) = 0 \implies k = j^{\frac{3}{2}} \implies k = (3\sqrt{3})^{\frac{3}{2}} (\dfrac{\sqrt{3} + 1}{\sqrt{3}}) \approx \boxed{18.683187}$ .

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