<s>[INST] Greatest volume of sphere inscribed in a cone with given Lateral area The Lateral area of a right circular cone is $S$. Fine the Greatest volume of Sphere that can be inscribed in it.
My try:
Given that $$\pi rl=S$$
Let the radius of the sphere inscribed in it be $R$
Let the semi vertex angle of cone be $\theta$
We have height of the cone as:
$$h=R+R\csc \theta$$
Now
$$\pi^2 r^2 l^2 =S^2$$ $\implies$
$$\pi^2 r^2 (r^2+h^2)=S^2$$
$$r^2(r^2+R^2(1+\csc \theta)^2)=\frac{S^2}{\pi^2}$$
So:
$$R^2=\frac{\frac{S^2}{\pi^2r^2}-r^2}{(1+\csc \theta)^2}\tag{1}$$
Also:
$$\sin \theta=\frac{r}{l}$$
So
$$\pi r \frac{r}{\sin \theta}=S$$
$\implies$
$$\csc \theta=\frac{S}{\pi r^2}$$
Hence $(1)$ becomes:
$$R^2=\frac{r^2(S^2-r^4\pi^2)}{(\pi r^2+S)^2}$$
$$R^2=r^2\frac{(S-\pi r^2)}{(S+\pi r^2)}$$
$$\pi R^2=\pi r^2\frac{(S-\pi r^2)}{(S+\pi r^2)}$$
Letting $\pi r^2=t$ we get
$$\pi R^2=f(t)=\frac{t(S-t)}{S+t}$$
Differentiating with respect to $t$ we get:
$$f'(t)=-1+\frac{2S^2}{(S+t)^2}=0$$ $\implies$
$$t=\left(\sqrt{2}-1\right)S$$
Hence
$$\pi r^2=\left(\sqrt{2}-1\right)S$$
$$r^2=\frac{\left(\sqrt{2}-1\right)S}{\pi}$$
So we get:
$$R^2=\frac{(3-2\sqrt{2})S}{\pi}$$
Hence Maximum volume of sphere is:
$$V=\frac{4\pi R^3}{3}=0.0534 (S)^{\frac{3}{2}}$$
But i feel this approach is tedious.Any better method?
 [/INST] The relation that you you desire for the radius of the incircle can found by examining the figure below. If $\theta$ is the half angle of the vertex, then clearly
$$
\sin\theta=\frac{r}{l}=\frac{R}{h-R}\\
h=\sqrt{l^2-r^2}
$$
From here you can find
$$R=\frac{rh}{l+r}=\frac{r\sqrt{l^2-r^2}}{l+r}=r\sqrt{\frac{l-r}{l+r}}$$
I believe this is what you are looking for.
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