Using Lagrange multiplier for n-polygon (Dido's problem)

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We have an example of the discrete version of Dido's problem.

Let $n\in N$ and $n \geq 3$.

A polygon exists with $n$ sides and corners P. The corners are given by:

$P_k$ = ($r_k \cos \sigma_k$, $r_k \sin \sigma_k$) with

$r_0$=$r_n$, $\sigma_0=0$, $\sigma_k$= $\alpha_1$ +...+$\alpha_k$, $k=1,...,n$ , $\alpha_j$ $\in$ ($0,\pi$), $j=1,...,n$ and $\sigma_n = 2\pi$.

See image https://postimg.cc/dhKtgwK3. (Credits to Amaan V)

We wrote the perimeter (length) and the area (A) as a function of the variables $r_n$ and $\alpha_n$.

length = $ \Sigma_{k=1}^n \sqrt {r_k^2+r_{k+1}^2-2r_k r_{k+1} \cos(\alpha_k)} $

and

A = $\Sigma_{k=1}^n \frac{1}{2} r_k r_{k+1} \sin(\alpha_k)$

For given, fixed perimeter (L) we are interested in the polygon with maximal area. We have to find a system of Lagrange equations for this problem. After that, we have to verify that the Lagrange equations are in the form $r_1 = ... = r_n$ and $\alpha_1 = ... = \alpha_n$ with corresponding Lagrange multiplier(s) in dependence on $n$.

Now I know how to use the Lagrange method for given equations, but for this discrete problem, I'm stuck. Please help me!

A Hint is given: A possible, straightforward formulation contains two Lagrangian multipliers and $2n + 2$ equations, including the restrictions.