Which $n$-gons, inscribed in a circle, have the largest area? Let $n$ be an integer, $n≥3$, and $H$ be a circle. Which (convex) $n$-gons, inscribed in $H$, have the largest area?
How can I use Lagrange multipliers to solve this?
 A: If you are forced to use Lagrange multipliers to solve this, you may do like this:
Let assume that $H$ has radius 1. Also, let $O$ be the center of $H$ and let $\theta_1, \cdots, \theta_n$ be central angles formed by two consecutive vertices of the $n$-gon and $O$.
Simple geometric argument shows that if this $n$-gon achieves its maximal area, then we must have $0 < \theta_i < \pi$. Thus we may assume this condition without loss of generality.
Now, the area $S(\theta_1, \cdots, \theta_n)$ of the $n$-gon is
$$ S(\theta_1, \cdots, \theta_n) = \frac{1}{2} \sum_{i=1}^{n} \sin \theta_i, $$
where $(\theta_1, \cdots, \theta_n)$ varies in the set
$$ D = \{ (\theta_1, \cdots, \theta_n) \in \Bbb{R}^n : \theta_i \in (0, \pi) \text{ and } \theta_1 + \cdots + \theta_n = 2\pi \}.$$
Applying Lagrange multiplier to $S$, there exists $\lambda$ such that
$$ \nabla S = \lambda \nabla (\theta_1 + \cdots + \theta_n) = (\lambda, \cdots, \lambda). $$
That is,
$$ \cos \theta_i = 2\lambda \quad \forall i. $$
But since $\theta_i \in (0, \pi)$, there is a unique $\theta_i$, say $\theta_i = \arccos(2\lambda)$, that satisfies this equation, and this shows that $\theta_1 = \cdots = \theta_n$. This proves that the $n$-gon with the maximal area must reduce to a regular $n$-gon.
