I have been battling with this for a little while now, but can't really get my head around it.
The question goes: how high must you pour water into a cone before it is half-full (by volume)?
No numbers specified, so I guess what I am looking for is a generalised formula for finding the height at which the volume is halved.
Edit: I was wrong in the last paragraph, what I was looking for was not half the volume, but the height at which the value of the volume was halved. Sorry.
Best Answer
The volume of a cone of base $R$ and height $H$ is
$$ V(R, H) = \frac{\pi}{3}R^2 H \tag{1} $$
You want to find a new base $r$ and height $h$ such such that
$$ V(r, h) = \frac{1}{2}V(R, H) \tag{2} $$
with the constraint
$$ \frac{R}{H} = \frac{r}{h} \tag{3} $$
Putting together (1), (2) and (3)
\begin{eqnarray} \frac{1}{2}\frac{\pi}{3}R^2 H &=& \frac{\pi}{3}r^2 h \\ \frac{1}{2}R^2 H &=& \left(\frac{R^2}{H^2}h^2 \right) h \\ \frac{1}{2}H &=& \frac{h^3}{H^2} \end{eqnarray}
From this you find
$$ h = 2^{-1/3}H $$