Sphere inside a cone!

By pythagoren theorem, we have $s=\sqrt{r^2+h^2}$ .

By similar triangles, we have

$\dfrac{0.5}{h-0.5}=\dfrac{r}{s}$ $\implies$ $\dfrac{0.5}{h-0.5}=\dfrac{r}{\sqrt{r^2+h^2}}$

Squaring both sides, we obtain

$\dfrac{0.25}{h^2-h+0.25}=\dfrac{r^2}{r^2+h^2}$

Cross-multiplying and simplifying, we get

$0.25r^2+0.25h^2=r^2h^2-r^2h+0.25r^2$

$0.25h^2=r^2h^2-r^2h$

$r^2=\dfrac{0.25h}{h-1}$

The volume of a cone is

$v=\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \left(\dfrac{0.25h}{h-1}\right)(h)=\dfrac{1}{3}\pi \left(\dfrac{0.25h^2}{h-1}\right)$

Differentiating both sides with respect to $h$ , we have

$\dfrac{dv}{dh}=\dfrac{1}{3}\pi \left[\dfrac{(h-1)(0.5h)-0.25h^2(1)}{(h-1)^2}\right]$

For minimum volume, $\dfrac{dv}{dh}=0$ , so we have

$(h-1)(0.5h)-0.25h^2=0$

$0.25h^2=0.5h$

$0.25h=0.5$

$h=\boxed{2~m}$

Let the base radius of the cone is r, and the height is h. So the volume V = πr^2 h/3. By similarity r/h = 0.5/√((h-0.5)^2-0.5^2). Square both side r^2/h^2 = 0.25/(h^2 – h),
So the volume V = 1/3 (π0.25) h^2/(h – 1), For minimum, dV/dh = 0., h=2 m