Circular Area maximising with polygons included. Help? given n concentric circles, each of which contain a point that can move on that circle, with radius’s {r1, r2, …., rn}, find the geometric(angular) arrangement of these points to get the polygon with maximum area.
the setup looks like the following:


the approach I used was to assign to each point a position vector to get a convenient means to calculate areas using cross-product. It ran into problems because of the following:
1) which vectors should be positioned together
2)the problem seems to be dependent on both, the number of point and their respective radiuses.
If there was some general method maybe using some inequality that assigns positions to the vectors or some other method that somehow automatically takes care of the nitty-gritty that would be great. 
I'm not exactly sure what to field to mark the problem in. All ideas and inputs and solutions are appreciated.
[I have a high-school math background]
 A: Here is an analysis using HwChu's suggestion of Lagrange multipliers.  As usual, Paul's Online Math Notes have a very accessible discussion of this subject.  Since the constraint is $\sum\theta_i=2\pi$, we have, in the notation of the notes, $$\begin{align}
f(\theta_1,\theta_2,\dots,\theta_n)&=\frac12\sum r_ir_{i+1}\sin\theta_i,\\
g(\theta_1,\theta_2,\dots,\theta_n)&=\sum\theta_i
\end{align}$$
(In the first equation, when $i=n$ we interpret $i+1$ as meaning $1$.)
We have $$\begin{align}
{\partial f\over\partial\theta_i}&=\frac12r_ir_{i+1}\cos\theta_i\\
\lambda{\partial g\over\partial\theta_i}&=\lambda
\end{align}
$$
So we see that $$\theta_i=\arccos\left(C\over r_ir_{i+1}\right)$$ for some constant $C$ such that $$\sum\theta_i = 2\pi$$
Except in the $n=4$ case, where we can take $C=0$, I think this will have to be solved numerically.
For example, for $n=5$ and radii $1,3,5,7,9$ WolframAlpha gives $C=2.53046$
EDIT
There's appears to be an oversight here.  Suppose we have the six radii $$1,3,5,7,9,15$$ so that the products of successive radii are $$ 3,15,35,63,135,15$$  The largest possible value of $C$ in the above is $3$, for otherwise $\arccos\left({C\over3}\right)$ is undefined.  Now$$ \cos^{-1}\left({3\over3}\right)+
\cos^{-1}\left({3\over15}\right)+
\cos^{-1}\left({3\over35}\right)+
\cos^{-1}\left({3\over63}\right)+
\cos^{-1}\left({3\over135}\right)+
\cos^{-1}\left({3\over15}\right)=7.295585>2\pi$$
Since $\cos^{-1}$ is a decreasing function, we see that the condition cannot be satisfied, which means that the maximum doesn't occur when the origin is inside the polygon.  This raises two questions.  First, when we do find a critical point by the above method, is it possible that it doesn't correspond to a global maximum, and second, how do we deal with the case where we know the maximum-area polygon doesn't contain the center?
