giúp em với ạ

Ta có: 
$\displaystyle h=\frac{72}{2x^{2}} =\frac{36}{x^{2}}$
→ $\displaystyle S_{tp} =2.2x^{2} +6x.\frac{36}{x^{2}}$
$\displaystyle =4x^{2} +\frac{216}{x}( x\neq 0)$
$\displaystyle S'_{tp} =8x-\frac{216}{x^{2}} =0$
$\displaystyle \Leftrightarrow 8x^{3} =216$
$\displaystyle \Leftrightarrow x=3$
BBT: 
$\displaystyle x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\infty \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\infty $
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
S'_{tp} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\ \ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ -\ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ +\\
S_{tp} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\infty \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\ \ \ +\infty \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\infty \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \searrow \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ \searrow \ \ \ \ \ \ \ \ \ \nearrow \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\infty \ ||\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 108\ 
\end{array}$
⟹ Diện tích toàn phần nhỏ nhất bằng $\displaystyle 108\left( cm^{2}\right)$