Okay so I was thinking about the following problem today: We have a guy who is h tall stand upon a paraboloid shaped hill of the form
$z=-ar^n$
How far away (in r) does his friend who is also h tall have to stand so that they just barely can't see each other (assuming they have eyes at h and nothing visible above it). This yields the equation for the line between their eyes (we can move to 2D cartesian):
$z = -ar_0^{n-1}r+h$
Where $r_0$ is the location of his friend. Solving these for the case where they have only one intersection yields:
$r_0=n^{\frac{1}{n-1}}\left(\frac{h}{a(n-1)}\right)^{1/n}$
Now an interesting question is whether there is some special hill "rank" such as a hill which minimizes the distance $r_0$. It is clear that $r_0$ tends to infinity as $n\rightarrow1$ so that is not interesting. Plotting $r_0$ as a function of n reveals that it indeed has a minimum but only if $a\gt h$.
Trying to to solve for this minimum seems impossible. The derivative with respect to n is really messy but finding it's root boils down to the equation:
$(n-1)^2\log\left(\frac{h}{a(n-1)}\right)+n^2\log(n) = 0$
Which I cant solve and Mathematica (at least with me using it) can't seem to solve it either.
Anything on how this might be solved?