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已知动点 ${\displaystyle P}$ 在圆 ${\displaystyle x^2+y^2=1}$ 上，定点 ${\displaystyle Q}$ 的坐标为 ${\displaystyle (1,1)}$, 动点 ${\displaystyle T}$ 满足 ${\displaystyle \overrightarrow{OT}=\lambda \overrightarrow{OP}+\mu \overrightarrow{OQ}}$, 其中 ${\displaystyle \lambda ,\mu \ge 0,{\lambda }^2+2{\mu }^2=4}$, 求 ${\displaystyle |QT|}$ 的最大值.

设 ${\displaystyle P(\cos \theta ,\sin \theta ),(\lambda ,\mu )=(2\cos \alpha ,\sqrt{2}\sin \alpha )}$, ${\displaystyle {\displaystyle \alpha \in [0,\frac{\pi }{2}]}}$, 则$$\begin{array}{r}\overrightarrow{OT}=(2\cos \alpha \cos \theta +\sqrt{2}\sin \alpha ,2\cos \alpha \sin \theta +\sqrt{2}\sin \alpha ),\end{array}$$
从而$$\begin{array}{rl}|QT|=& \sqrt{(2\cos \alpha \cos \theta +\sqrt{2}\sin \alpha -1{)}^2+(2\cos \alpha \sin \theta +\sqrt{2}\sin \alpha -1{)}^2}\\ =& \sqrt{2(\sqrt{2}\sin \alpha -1{)}^2+4{\cos }^2\alpha +4(\sqrt{2}\sin \alpha -1)\cos \alpha (\sin \theta +\cos \theta )}\\ \le & \sqrt{6-4\sqrt{2}\sin \alpha +4\sqrt{2}|(\sqrt{2}\sin \alpha -1)\cos \alpha |},\end{array}$$

• 若 ${\displaystyle \alpha \in {\displaystyle [\frac{\pi }{4},\frac{\pi }{2}]}}$, 记 ${\displaystyle \sin \alpha +\cos \alpha =t\in [1,\sqrt{2}]}$, 则 ${\displaystyle 2\sin \alpha \cos \alpha =t^2-1}$, 从而$$\begin{array}{r}|QT|\le \sqrt{4(t^2-1)-4\sqrt{2}t+6}\le \sqrt{2},\end{array}$$
• 若 ${\displaystyle \alpha \in {\displaystyle [0,\frac{\pi }{4}]}}$, 记 ${\displaystyle \cos \alpha -\sin \alpha =s\in [0,1]}$, 则 ${\displaystyle 2\sin \alpha \cos \alpha =1-s^2}$, 从而$$\begin{array}{r}|QT|\le \sqrt{6+4\sqrt{2}s-4(1-s^2)}\le 2+\sqrt{2},\end{array}$$

从而 ${\displaystyle |QT{|}_{max}=2+\sqrt{2}}$. 取等条件是 ${\displaystyle {\displaystyle P(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}),(\lambda ,\mu )=(2,0)}}$.

可以进行简化计算$$\begin{array}{r}|\overrightarrow{QT}|=|(\mu -1)\overrightarrow{OQ}+\lambda \overrightarrow{OP}|\le \sqrt{2}|1-\mu |+\lambda ,\end{array}$$
而 ${\displaystyle 0\le \lambda \le 2,0\le \mu \le {\displaystyle \frac{\sqrt{2}}{2}\Rightarrow |1-\mu |\le 1}}$, 从而立得 ${\displaystyle |QT|\le \sqrt{2}+2}$.