= Solution
If $z(t)$ is the water depth and $A(z)=\pi r(z)^2$ is the horizontal cross-sectional area, <Torricelli's law> gives the volume flux through the hole as
$$
q_0=\pi r_0^2\sqrt{2gz}.
$$
Conservation of volume gives
$$
A(z)\dot z=-\pi r_0^2\sqrt{2gz}.
$$
For a prescribed constant fall rate $\dot z=-\alpha$, $\alpha>0$, this requires
$$
A(z)=\frac{\pi r_0^2\sqrt{2g}}{\alpha}\sqrt z.
$$
Consequently
$$
\boxed{r(z)=
r_0\left(\frac{\sqrt{2g}}{\alpha}\right)^{1/2}z^{1/4}},
$$
so the container radius must be proportional to the fourth root of height.
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