11. Inside a cone, there is a sphere that is tangent to the cone's side and base. The maximum ratio of the sphere's surface area to the cone's surface area is $\qquad$ .
Let the radius of the base of the cone be $R$, and the slant height be $l$, then the radius of the inscribed sphere $r=\frac{R \cdot \sqrt{l^{2}-R^{2}}}{l+R}$.
Thus, $\frac{S_{\text {sphere }}}{S_{\text {cone }}}=\frac{4 \pi r^{2}}{\pi R l+\pi R^{2}}=\frac{\frac{4 R^{2}\left(l^{2}-R^{2}\right)}{(l+R)^{2}}}{R(l+R)}=\frac{4 R(l-R)}{(l+R)^{2}} \leqslant \frac{2 \cdot\left(\frac{2 R+l-R}{2}\right)^{2}}{(l+R)^{2}}=\frac{1}{2}$, equality holds when $2 R=l-R \Rightarrow l=3 R$.