Maximum moment of inertia of arc Is there is variational calculus solution to the problem of maximum moment of inertia of a wire of uniform density per unit length $s$ between two fixed endpoints, about z-axis in 3-Space?
Considering cylindrical coordinates:  $\int ( \sqrt{( r^2+r^{'2}+z^{'2})} - \lambda ) ds $ may need to be maximized.
 A: Without further constraints, there is no maximum.  You can always increase the moment of inertia by putting more mass farther away from the z-axis.  However, if you constrain (say) the length of the wire, there may be a maximum, and there may also be a minimum (even unconstrained)...
Here's an overview of the simplest variational problem:
The moment of inertia is given by $\rho \int_c r^2 \sqrt{\dot{r}^2+(r\dot{\theta})^2+\dot{z}^2} dt = \rho \int_c (x^2+y^2) \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} dt$, where $\rho$ is the wire's density, $r=\sqrt{x^2+y^2}$ is the radial distance from the z-axis, and the arc c is parametrized by t.  A stationary point must be stationary under variations of x, y, and z independently, and thus we get an Euler-Lagrange equation for each coordinate:
$$
\frac{d}{dt}\left( \frac{(x^2+y^2)\dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}}\right) - 2x\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}=0 \\
\frac{d}{dt}\left( \frac{(x^2+y^2)\dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}}\right) - 2y\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}=0 \\
\frac{d}{dt}\left( \frac{(x^2+y^2)\dot{z}}{\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}}\right)=0
$$
or in polar coordinates:
$$
\frac{d}{dt}\left( \frac{r^2\dot{r}}{\sqrt{\dot{r}^2+(r\dot{\theta})^2+\dot{z}^2}}\right) - 2r\sqrt{\dot{r}^2+(r\dot{\theta})^2+\dot{z}^2}-\frac{r^3\dot{\theta}^2}{\sqrt{\dot{r}^2+(r\dot{\theta})^2+\dot{z}^2}}=0 \\
\frac{d}{dt}\left( \frac{r^4\dot{\theta}}{\sqrt{\dot{r}^2+(r\dot{\theta})^2+\dot{z}^2}}\right)=0 \\
\frac{d}{dt}\left( \frac{r^2\dot{z}}{\sqrt{\dot{r}^2+(r\dot{\theta})^2+\dot{z}^2}}\right)=0
$$
I won't try to solve these here. 
[PS: I would like to say that we can choose to parametrize by arc length so that the square root in the above expressions goes away, because then the equations would reduce to:
$$
r^4 \dot{\theta} = c_1 \\
r^2 \dot{z} = c_2 \\
\frac{d}{dt} \left(r^2\dot{r}\right) - 2r - \frac{c_1^2}{r^5}=0
$$
However, parametrizing by arc length will cause the bounds of the moment of inertia integral to change with variations to the curve, which complicates the variational problem. ]
[PSPS: If you want to look into constraints on total arc length, you'll want to add a Lagrange multiplier.  Check out the book "Mathematics of Classical and Quantum Physics" by Byron & Fuller for details. ]
