Jtdylktuy

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 = $2sin alpha cos alpha$

and $h$ = height of the cone = $2 (cos alpha )^2$.

Then the surface of the cone is:
$$S(alpha)=4 pi sin alpha (cos alpha)^2+4 pi (sin alpha)^2 (cos alpha)^2=4 pi (sin alpha (cos alpha)^2+(sin alpha)^2 (cos alpha)^2)=
4 pi (-(sin alpha)^4-(sin alpha)^3+(sin alpha)^2+sin alpha)$$

$$S(alpha)'=cos alpha (-4 (sin alpha)^3-3 (sin alpha)^2 +2sin alpha +1)=0 iff alpha=0 lor sinalpha approx 0,64 $$

$0$ 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 $S'(alpha)=0$. $$begin{aligned}S'(alpha)&=cos alpha (-4 (sin alpha)^3-3 (sin alpha)^2 +2sin alpha +1)\&=cosalpha(sinalpha+1)(1+sinalpha-4sin^2alpha)end{aligned}$$
Thus, equation $S'(alpha)=0$ is equal to $cosalpha=0$ or $sinalpha=-1$ or $1+sinalpha-4sin^2alpha=0$.

First equation has solution in region $left[0;fracpi2right]$ and is equal to $fracpi2$. The second equation hasn't any solution in permissible region. The third equation can be solved by substitution $t=sinalpha$. So we get $$1+t-4t^2=0$$ and $$t_{1,2}=frac{1pmsqrt{17}}8$$

Since $0leqalphaleqfracpi2$, $0leqsinalphaleq1$ and $t_1=frac{1-sqrt{17}}8$ is out of permissible region. Thus, $alpha=fracpi2$ and $alpha=arcsinfrac{1+sqrt{17}}8$ are the only solutions of $S'(alpha)=0$.

Finally we get $S(fracpi2)=0$ and $Sleft(arcsinfrac{1+sqrt{17}}8right)=frac{pi(107+51 sqrt{17})}{128}
$.