We are at the space $R^n$ with $3 \le n$ . Let $S$ be the unit circle. I want to find the convex polygon with the maximal area, when the polygon has $n$ vertices and each one is in $S$.
It looks like classic to solve it with the Lagrange multipliers since there is a term that has to be satisfied and we are also looking for critical point that satisfies that term.
However, I don't really understand what function gives the area of the polygon, and without it I can't really continue.
Lets consider $\max\limits_P \mathrm{Area}(P)$ where the $\max$ ranges over all polygons $P\subset \bar B(0,1)$ with at most $n$ vertices.
First of all, the polygon can be assumed to be convex, because taking the convex hull reduces the number of vertices while increasing the area.
Second, the vertices can be assumed to lie on the circumference, because if one of them does not, you can move it orthogonally to the segment joining the two adjacent vertices until it touches the circle and the area will increase.
So we have vertices $v_1,\dots,v_n\in S^1$. Let's call $\alpha_1,\dots,\alpha_n$ the angles between them. We can also assume that all the angles are $\alpha_i\leq\pi$, otherwise the configuration is easy to improve by hand once again.
Then $\alpha_1+\dots+\alpha_n=2\pi$ and $$ \mathrm{Area}(P) = \sum_{i=1}^n \sin(\alpha_i/2)\cos(\alpha_i/2) = \frac12 \sum_{i=1}^n \sin(\alpha_i). $$
With Lagrange multipliers you get that $$ \bigl(\cos(\alpha_1),\dots,\cos(\alpha_n)\bigr) = \lambda (1,\dots,1), $$ so the angles have to be all equal, hence the polygon is regular.
Of course, there are easier ways to maximize the last sum: for example, you can substitute two consecutive angles $\alpha_i,\alpha_{i+1}$ with their average $\tfrac{\alpha_i+\alpha_{i+1}}2$ and show that the area increases; therefore the optimal polygon must have all equal angles.