3. The length of the diagonal of a rectangular parallelepiped is 3. What is the maximum possible value of the surface area of such a parallelepiped?
Answer: 18.

Solution. Let the edges of the rectangular parallelepiped be $-a, b$, and $c$. According to the problem, its diagonal is $\sqrt{a^{2}+b^{2}+c^{2}}=3$, and thus, $a^{2}+b^{2}+c^{2}=9$. The surface area of the parallelepiped is $f=2(a b+b c+c a)$. We will prove the inequality $a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}$. Indeed, multiply both sides by 2 and write it in an equivalent form:

$$
2 a b+2 b c+2 c a \leqslant 2 a^{2}+2 b^{2}+2 c^{2} \quad \Longleftrightarrow \quad(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \leqslant 0
$$

From the proven inequality, it follows that the maximum value of the function $f$ does not exceed $2 \cdot 9=18$, and the equality sign is possible only for a rectangular parallelepiped where all edges are equal, $a=b=c=\sqrt{3}$, i.e., for a cube with edge $\sqrt{3}$.

Criteria. Correct answer without explanation - 0 points. Correctly formulated function $f-1$ point. Noted (without justification) that the maximum value is achieved for a cube and correctly calculated the length of its edge - another 1 point. Proof of the estimate without specifying an example - 5 points. Complete solution - 7 points.