Belorusija, 2012

Answer: $12$.
We use the following obvious inequalities. If the quadrilateral $ABCD$ has the sides $AB = a$, $BC = b$, $CD = c$, $DA = d$, and the area $S$, then
$$S \le \frac{ab + cd}{2} \text{ because}$$
$$S = S(ABC) + S(ACD) = \\ = \frac{1}{2}ab \sin \beta + \frac{1}{2}cd \sin \delta \le \frac{ab + cd}{2}.$$
Similarly, $S \le \frac{ad + bc}{2}$.
Summing these inequalities, we easily obtain $S \le \frac{(a+c)(b+d)}{4}$. Hence
$$S \le \frac{6 \cdot 8}{4} = 12.$$ It is easy to see that equality occurs for a rectangle.