Gọi R là bán kính đáy, l là đường sinh. Chiều cao khối nón là SH = h.

Ta có: ${{S}_{tp}}={{S}_{xq}}+{{S}_{d}}\Leftrightarrow \pi =\pi Rl+\pi {{R}^{2}}$

$\Leftrightarrow 1=R\sqrt{{{h}^{2}}+{{R}^{2}}}+{{R}^{2}}$

$\Leftrightarrow \left\{ \begin{align} & 1-{{R}^{2}}=R\sqrt{{{h}^{2}}+{{R}^{2}}} \\ & R\le 1 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align} & 1-2{{R}^{2}}={{R}^{2}}{{h}^{2}} \\ & R\le 1 \\ \end{align} \right.$ $\Leftrightarrow \left\{ \begin{align} & R\le 1 \\ & {{R}^{2}}=\frac{1}{2+{{h}^{2}}} \\ \end{align} \right.$

Ta có: $V=\frac{1}{2}\pi {{R}^{2}}h=\frac{1}{3}\pi \frac{h}{{{h}^{2}}+2}\le \frac{1}{3}\pi \frac{h}{2\sqrt{2}h}=\frac{\pi }{6\sqrt{2}},$do đó

${{V}_{\max }}=\frac{\pi }{6\sqrt{2}}\Leftrightarrow \left\{ \begin{align} & {{h}^{2}}=2 \\ & {{R}^{2}}=\frac{1}{{{h}^{2}}+2} \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & h=\sqrt{2} \\ & R=\frac{1}{2} \\ \end{align} \right.$