Working out the triangle length and prism length of a equilateral triangular prism?

$$Let the triangle have side length 2a, the perpendicular height will be $a\sqrt3$\\ Let the depth of prism be D (all units in cm)\\ A $1cm^3$ container has the capacity of $1mL$\\ Volume of prism = Area of triangle $\times$ Depth\\ $5178 = \frac{1}{2} \times 2a \times a\sqrt3 \times D$\\ $5178 = a^2\sqrt3 \times D$\\\\ $D=\dfrac{5178}{a^2\sqrt3} $\\$$

$$$D=\dfrac{5178}{a^2\sqrt3} $\\\\ Surface area\\ S=2(triangles)+3(rectangles)\\\\ $S=2(\frac{1}{2}\times 2a \times a\sqrt3)+3(2a\times D)$\\\\ $S=2(a^2\sqrt3)+3(2a\times \frac{5178}{a^2\sqrt3})$\\\\ $S=2\sqrt3a^2+\frac{5178*6}{a\sqrt3}$\\\\ $S=2\sqrt3a^2+\frac{31068a^{-1}}{\sqrt3}$\\\\$$

$$Surface area will be minimum when \\\\ $\frac{dS}{da}=0\;\;and\;\;\frac{d^2S}{da^2}>0$\\\\ $\frac{dS}{da}=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\ $0=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\ $0=12a-31068a^{-2}$\\\\ $0=12a^3-31068$\\\\ $12a^3=31068$\\\\ $a^3=2589$\\\\ $a\approx 13.73$cm\\\\\\ $\frac{d^2S}{da^2}=4\sqrt3+\frac{2*31068a^{-3}}{\sqrt3}>0\;for\; all \;a>0$\\\\ Therefore any turning point must be a minimum.$$

$$\\a\approx 13.73cm\\\\ D\approx \frac{5178}{13.73^2*\sqrt3}\\\\ D\approx 15.86cm\\\\\\ \mbox{The triangle has side length 2*13.73= 27.46cm and the prism is 15.86cm deep}\\\\ check\\ V=1/2*27.46*13.73*\sqrt3*15.86 = 5178.5cm^3 \qquad good$$  

I think that is okay