maximal surface of a cone inscribed in a sphere with radius 1 Amongst all the cones inscribed in a sphere of radius 1, I have to find that one of maximal surface.
Every cone is characterized by a value for α, the opening angle of the cone, 
 with $\alpha \in [0, \frac{\pi}{2}]$ (the angle between the vertical line passing through the centre of the sphere and the apotema con the cone)
Applying some properties of the rectangular triangles:
$r$ = radius of the base = $2\sin \alpha \cos \alpha$
and $h$ = height of the cone = $2 (\cos \alpha )^2$.
Then the surface of the cone is:
$$S(\alpha)=4 \pi \sin \alpha (\cos \alpha)^2+4 \pi (\sin \alpha)^2 (\cos \alpha)^2=4 \pi (\sin \alpha (\cos \alpha)^2+(\sin \alpha)^2 (\cos \alpha)^2)=
4 \pi (-(\sin \alpha)^4-(\sin \alpha)^3+(\sin \alpha)^2+\sin \alpha)$$
$$S(\alpha)'=\cos \alpha (-4  (\sin \alpha)^3-3 (\sin \alpha)^2 +2\sin \alpha +1)=0      \iff \alpha=0 \lor \sin\alpha \approx 0,64 $$
$0$    is the minimum solution , while $0,64$ is our solution.
Then $S_{max}=4 \pi (-0,64^4+0,64^3+0,64^2+0,64) =14,37$
In the book the suggested solution is $\frac{\pi(107+51 \sqrt{17})}{128}$.
Can someone help me to understand how did it find it?

I'll start from the equation $S'(\alpha)=0$. $$\begin{aligned}S'(\alpha)&=\cos \alpha (-4  (\sin \alpha)^3-3 (\sin \alpha)^2 +2\sin \alpha +1)\\&=\cos\alpha(\sin\alpha+1)(1+\sin\alpha-4\sin^2\alpha)\end{aligned}$$
Thus, equation $S'(\alpha)=0$ is equal to $\cos\alpha=0$ or $\sin\alpha=-1$ or $1+\sin\alpha-4\sin^2\alpha=0$. 
First equation has solution in region $\left[0;\frac\pi2\right]$ and is equal to $\frac\pi2$. The second equation hasn't any solution in permissible region. The third equation can be solved by substitution $t=\sin\alpha$. So we get $$1+t-4t^2=0$$ and $$t_{1,2}=\frac{1\pm\sqrt{17}}8$$
Since $0\leq\alpha\leq\frac\pi2$, $0\leq\sin\alpha\leq1$ and $t_1=\frac{1-\sqrt{17}}8$ is out of permissible region. Thus, $\alpha=\frac\pi2$ and $\alpha=\arcsin\frac{1+\sqrt{17}}8$ are the only solutions of $S'(\alpha)=0$.
Finally we get $S(\frac\pi2)=0$ and $S\left(\arcsin\frac{1+\sqrt{17}}8\right)=\frac{\pi(107+51 \sqrt{17})}{128}
$.
