20. Let $a$ and $b$ be real numbers such that $17\left(a^{2}+b^{2}\right)-30 a b-16=0$. Find the maximum value of $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$.
(2 marks)
設 $a$ 、 $b$ 為實數 , 使得 $17\left(a^{2}+b^{2}\right)-30 a b-16=0$ 。 求 $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$ 的最大值。
20. 7
20. Rewrite the given equation as $(a+b)^{2}+16(a-b)^{2}=16$. Hence we may let $a+b=4 \cos x$ and $a-b=\sin x$. Note that
$$
\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}=\sqrt{(4 a-2 b)^{2}-3(4 a-2 b)+9}=\sqrt{\left(4 a-2 b-\frac{3}{2}\right)^{2}+\frac{27}{4}} .
$$

Since $4 a-2 b=(a+b)+3(a-b)=4 \cos x+3 \sin x$, whose value lies between -5 and 5 , the maximum value of the above expression occurs when $4 a-2 b=-5$, and the maximum value is $\sqrt{\left(-5-\frac{3}{2}\right)^{2}+\frac{27}{4}}=7$. (The corresponding values of $a$ and $b$ can be found by solving the equations $4 a-2 b=-5$ and $(a+b)^{2}+16(a-b)^{2}=16$, giving $a=-\frac{19}{10}$ and $b=-\frac{13}{10}$.)