Lượng giấy tiêu thụ ít nhất S_xq nhỏ nhất

Ta có $\left\{ \begin{align}  & V=27=\frac{1}{3}\pi {{r}^{2}}h\Rightarrow h=\frac{81}{\pi {{r}^{2}}} \\  & {{S}_{xq}}=\pi rl=\pi r\sqrt{{{h}^{2}}+{{r}^{2}}} \\ \end{align} \right.\Rightarrow {{S}_{xq}}=\pi r\sqrt{{{\left( \frac{81}{\pi {{r}^{2}}} \right)}^{2}}+{{r}^{2}}}=\sqrt{\frac{{{81}^{2}}}{{{r}^{2}}}+{{\pi }^{2}}{{r}^{4}}}$

Lại có $\frac{{{81}^{2}}}{{{r}^{2}}}+{{\pi }^{2}}{{r}^{4}}=\frac{{{81}^{2}}}{2{{r}^{2}}}+\frac{{{81}^{2}}}{2{{r}^{2}}}+{{\pi }^{2}}{{r}^{4}}\ge 3\sqrt[3]{\frac{{{81}^{2}}}{2{{r}^{2}}}.\frac{{{81}^{2}}}{2{{r}^{2}}}.\pi {{r}^{4}}}=3\sqrt[3]{\frac{{{81}^{2}}}{2{{r}^{2}}}.\frac{{{81}^{2}}}{2{{r}^{2}}}.\pi {{r}^{4}}}$không đổi

Dâu “=” xảy ra ó $\frac{{{81}^{2}}}{2{{r}^{2}}}=\frac{{{81}^{2}}}{2{{r}^{2}}}={{\pi }^{2}}{{r}^{4}}\Leftrightarrow {{r}^{6}}=\frac{{{81}^{2}}}{2{{\pi }^{2}}}\Leftrightarrow r=\sqrt[6]{\frac{{{81}^{2}}}{2{{\pi }^{2}}}}=\sqrt[6]{\frac{{{3}^{8}}}{2{{\pi }^{2}}}}$