Belarusian Mathematical Olympiad

Answer: $S(ABCD) = \frac{L^2}{8}$.
Let the area of $ABCD$ be a maximum for some $AB = x$, $BD = y$, $CD = z$, $x + y + z = L$. Since $S(ABCD) = S(ABD) + S(DBC) = \frac{1}{2} AB \cdot BD \sin \angle ABD + \frac{1}{2} BD \cdot BC \sin \angle BDC$, we see that the area of the quadrilateral with fixed values of $x$, $y$, $z$ is maximum if $\angle ABD = \angle CDB = 90^\circ$. Therefore,
$$S(ABCD) = \frac{1}{2} x y + \frac{1}{2} y z = \frac{1}{2} y(x+z) = \frac{1}{2} y(L-y).$$
It is easy to see that for $0 < y < L$ the value of the product $y(L-y)$ is a maximum if $y = \frac{L}{2}$ and it is equal to $\frac{L^2}{4}$. Therefore, the maximal value of $ABCD$ with given sum of its sides $AB$, $CD$ and diagonal $BD$, $AB + BD + DC = L$, is equal to $\frac{L^2}{8}$.