In this post, it is asked how to maximize the area of a given quadrilateral, with a given angle $\alpha$ and perimeter $N$. I proposed the following approach.
Consider the quadrilateral $Q$ in the image below :
The area of $Q$ is given by $$ f(a,b,c,d,A,C) = \frac{a d \sin A+bc \sin C}{2} $$
So here we want to maximize $f(a,b,c,d,A,C)$ subject to : $$ A = \alpha \\ a+b+c+d = N \\ $$
That is, we want to maximize $$ f(a,b,c,C) = \frac{a (N-a-b-c) \sin \alpha +bc\sin C}{2} $$
Assuming $\alpha,b,c >0$, solving $\nabla f = 0$ yields : $$ C = \frac{\pi}{2}\\ a = \frac{N}{2}\frac{1}{1+\sin \alpha} \\ b=c= \frac{N}{2}\frac{\sin \alpha}{1+\sin \alpha} $$ If we can show that $f$ is concave on this point then we could conclude that the maximum area equals $$ \boxed{ A(\alpha,N) = \frac{1}{8}\frac{N^2 \sin \alpha }{\sin \alpha+1} } $$
But this is wrong, as shown by simulations in the post. For example $$ A(\frac{\pi}{6},60) = 150 $$ while setting $a$ to $22.78$, $b$ and $c$ to $7.22$ and $C$ to $110^°$ yields an area of approximately $154.3 > 150$.
My question : Where is the flaw in this answer ? The closed form of the area seems to be correct for the particular case of a square ($A(\frac{\pi}{2},N) = \frac{N^2}{16}$).
You may miss an additional constraint that two triangles share the side BD, so at least we should make sure $$ a^2 + d^2 - 2ad \cos A = b^2 + c^2 - 2bc \cos C.$$