, 2012

Answer: $S(ABCD) = L^2/8$.
Let the area of $ABCD$ be a maximum for some $AB = x$, $BD = y$, $CD = z$, $x + y + z = L$. Since $S(ABCD) = S(ABD) + S(DBC) = \frac{1}{2} AB \cdot BD \sin \angle ABD + \frac{1}{2} BD \cdot DC \sin \angle BDC$, 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,
$$\begin{aligned} S(ABCD) &= \frac{1}{2}xy + \frac{1}{2}yz = \frac{1}{2}y(x+z) = \\ &= [x+y+z = L \Rightarrow x+z = L-y] = \frac{1}{2}y(L-y). \end{aligned}$$

It is easy to see that for $0 < y < L$ the value of the product $y(L-y)$ is a maximum if $y = L/2$ and it is equal to $L^2/4$. Therefore, the maximal value of $ABCD$ with given sum of its sides $AB$, $CD$ and diagonal $BD$, $AB + BD + DC = L$, is equal to $L^2/8$.