Find the quadrilateral of maximal perimeter of fixed area $15$ enclosed in a square of side $4$.
One (trivial) remark: we cannot take a concave quadrilateral because any concave quadrilateral is contained in a triangle, which must in turn be contained in the square by convexity, and a triangle contained in a square of side $4$ has maximum area $8<15$. Therefore the quadrilateral must be convex. Hence the perimeter cannot be larger than that of the square ($16$). Also, we might thus decompose the quadrilateral as the union of $4$ triangles by drawing the diagonals.
See the square as $[-2,2]^2\subset \mathbb{R}^2$.
My idea: the sought quadrilateral is a rhombus having two vertices $A=(-2,-2)$, $C=(2,2)$ (on one diagonal of the square) and the other two $D=(a,-a)$ and $B=(-a,a)$ (on the other diagonal of the square, equally distant from the center), with the parameter $a\in (0,2)$ chosen so that the area is actually $15$ (actually $a=15/8$, which yields the perimeter $\sqrt{1+31^2}/2\approx 15.51$).
I would say that because without the condition of being enclosed in a square, we can arbitrarily increase the perimeter of a quadrilateral with a given area by making the angles on two opposite vertices $A,C$ vanish (while the others tend to $\pi$), and their distance increase to infinity. Inside the square, we can maximize the distance between $A$ and $C$ by taking them to be opposite vertices of the square, i.e. $A=(-2,-2)$ and $C=(2,2)$.
With this assumption, the situation is simpler. But I still need another assumption: the two triangles $ABC$ and $ACD$ have same area. Then for the area to be fixed, $B$ and $D$ must lie on a specific line parallel to the diagonal. At this point, it is easy (one-variable calculus) to show that the perimeter of $ABC$ is maximized if $B$ lies on the other diagonal of the square, and the same goes for $D$.
EDIT: This is wrong and lead me to wrong conclusions. The isosceles triangle minimizes, not maximizes, the perimeter! As a result, the point on the line which maximizes the perimeter is the furthest away point of intersection with the border of the square. In particular, all vertices of the quadrilateral must lie on the border of the square.
Since the areas must be the same, the distances of $B$ and $D$ from the center must also be the same.
Either way, I have no idea how to make my assumptions more formal. I'm not even sure that my solution is correct.