What is the maximum area of a polygon with sides of lengths $1,2,3,\ldots,N$?
Intuition tells me the polygon must be inscribed in a regular polygon with $1+2+3+ \cdots +N$ sides.
What would be the limit of the surface area of a polygon with $N \to \infty$ if the series converge?
As I mentioned in the comments, you just need to place the vertices on a circle. So if the radius $r$ is known, then the area is the sum of the areas of a bunch of isosceles triangles with base length $1,2,3,...,n$ and slanted length $r$. So they have height $\sqrt{r^2-1/2},\sqrt{r^2-2/2},...,\sqrt{r^2-n/2}$. So the total area is $1\sqrt{r^2-1/2}+2\sqrt{r^2-2/2}+...+n\sqrt{r^2-n/2}$.
To find the radius $r$ note that the sum of the angles of the isosceles triangles must be $2\pi$. So we have $\sin^{-1}(\frac{1/2}{r})+\sin^{-1}(\frac{2/2}{r})+...+\sin^{-1}(\frac{n/2}{r})=2\pi$. I don't know if there is an analytic or geometric way to solve for $r$, but computationally you can simply do a binary search.
EDIT: By Brahmagupta's formula at least for $n=4$ we can calculate the maximal area is $\sqrt{(s-1)(s-2)(s-3)(s-4)}$ where $s=(1+2+3+4)/2=5$, so the area is $\sqrt{4!}=2\sqrt{6}\approx4.9$.
EDIT2: Apparently there are formulas like Brahmagupta's for up to $8$ sides. However, more than $8$ sides appears to be a currently unsolved problem. Hence, I expect you will not get a better answer than binary search.
However, as I mentioned in the comments $\lim_{n\to\infty}A(n)/(S(n))^2=1/\pi$, so you can use this to approximate $A(n)$. However, binary search definitely allows more precise approximations.