6. Let $x$ and $y$ be non-negative integers such that $69 x+54 y \leq 2008$. Find the greatest possible value of $x y$.
(1 mark)
設 $x$ 、 $y$ 爲滿足 $69 x+54 y \leq 2008$ 的非負整數。求 $x y$ 的最大可能值。
(1 分)
6. 270
6. Note that for non-negative real numbers $a$ and $b$, we have $(\sqrt{a}-\sqrt{b})^{2} \geq 0$ and hence $a+b \geq 2 \sqrt{a b}$. It follows that $2008 \geq 69 x+54 y \geq 2 \sqrt{69 x \cdot 54 y}$ and hence
$$
x y \leq\left(\frac{2008}{2 \sqrt{69 \cdot 54}}\right)^{2}=\frac{1004^{2}}{69 \cdot 54} \approx 270.5 .
$$

Thus $x y$ is at most 270. This maximum is attainable when $x=15$ and $y=18$, as $69(15)+54(18)=2007$. Therefore the answer is 270 .