12 Find the maximum value of $\sqrt{x-144}+\sqrt{722-x}$.
12. Answer: 34
From AM-GM: $\sqrt{a b} \leq \frac{a+b}{2}$, we have $\sqrt{a}+\sqrt{b} \leq \sqrt{2(a+b)}$. Hence $\sqrt{x-144}$ $+\sqrt{722-x} \leq \sqrt{2(722-144)}=34$.