I found this interesting problem in Introduction to Classical Mechanics with Problems and Solutions by David Morin:
Given a point $P$ in space, and given a piece of malleable material of constant density, how should you shape and place the material in order to create the largest possible gravitational field at $P$?
Any ideas?
This Physics quiz website by Yacov Kantor provides the solution in the February 2002 quiz. The optimal surface profile (with max gravity in the origin) in spherical and cylindrical coordinates for the solid of revolution is $r^2=z_0^2 \cos\theta$ and $(z^2+\rho^2)^{3/2}= z_0^2z$, respectively, $0\leq z\leq z_0$. The gravity in the origin is only 2.6% larger than the gravity on the surface of a spherical planet.
My first guess is to shape the material into a unit-radius sphere and place it so that P is on its surface- this way, you don't 'lose' any gravitational field due either to the 1/r^2 term, or to any asymmetric considerations in the direction of the vector pointing from the source.
In other words- here we have a maximally symmetric source, and we are sitting right on top of it. I'd be interested to see if anything more exotic comes up, though!
Hmmm...that's fun.
It is important to note that we seek the largest possible field, and not the lowest potential versus infinity which might yield a lower result.
So let's consider some obvious cases (all of which I will reduce to a dependence only on $\rho$ and $M$.).
A nearby sphere
We form the material into a sphere of volume $V=M/\rho$ and radius $R = 3/4 V^{1/3}/\pi$, place it next to $p$ and compute the field as $F_G = G M/R^2 = (3/4) G M V^{-2/3} = (3/4) G M^{1/3} \rho^{-2/3}$
A nearby plane segment
We form the material into a disk of thickness $T$ radius $R = 10 t$ (claiming that this is sufficient for a "infinite place" approximation near the center), so that $V = 100 t^3$so that $t = (V/100)^{1/3} = [M/(100 \rho)]^{1/3}$. We place the it such that $p$ borders center on one side. The surface density is $\sigma = \rho * t = 100^{-2/3} \rho^{2/3} M^{1/3}$ and the field is $F_G = 2 \pi G \sigma = 2 \pi 100^{-2/3} G M^{1/3} \rho^{-2/3}$.
Note that so far, both distributions result in a form $F_G = C \times G M^{1/3} \rho^{-2/3}$, so we need only compare the constants. Well, $2 \pi 100^{-2/3} = 2 \pi (0.046) = 0.29$, so the spherical distribution is better than the flat one.
The trivial solution is compact the mass to a point object, I guess. In that case the gravity at point P is infinite, because the distance is zero. I'm not sure if this violates the 'constant density' principle though.
I figured out a solution using calculus of variations. From the context of the problem, this given material should also have constant mass, and because the material also has constant density, this implies constant volume. Now we can argue using symmetry that the overall shape that gives maximum gravitational field at the origin must be symmetric about some axis, so let's choose the $z$ axis for simplicity. Let's do the rest of this problem in spherical coordinates. The $z$ component of the gravitational field is then the volume integral of $\frac{G\ \mathrm{d}m\ \cos(\theta)}{r^2}$, and $\mathrm{d}m=\mathrm{density}\times r^2\sin(\theta)\ \mathrm{d}r\ \mathrm{d}\theta\ \mathrm{d}\phi$. Using the symmetry argument, you can reduce the volume integral into an integral of a single variable WRT $\theta$. So, we have what we need to answer the question. It's basically an optimization of $g_z$, the $z$ component of $g$, with the constraint that the volume is held fixed.
