每日一题_191214

已知$O$为坐标原点,圆$M:(x+1)^2+y^2=1$,圆$N:(x-2)^2+y^2=4$.$A,B$分别为圆$M$和圆$N$上的动点,则$S_{\triangle OAB}$的最大值为$\underline{\qquad\qquad}$.
解析:
法一 如图,若固定$A$的位置,则当$B$位于如图所示位置时,$S_{\triangle OAB}$的面积最大,

若设$\angle CON=\theta$,根据对称性,仅需考察$\theta\in\left[0,\dfrac{\pi}{2}\right)$的情形,此时 $$\begin{split} &S_{\triangle OAB}\ =&\dfrac{1}{2}\cdot |OA|\cdot |BC|\ =&\dfrac{1}{2}\cdot |OC|\cdot \left(|NC|+|NB|\right)\ =&\dfrac{1}{2}\cdot |ON|\cos\theta\cdot\left( |ON|\sin\theta+|NB|\right)\ =&2\cos\theta\left(1+\sin\theta\right)\ =&2\sqrt{\cos^2\theta\cdot\left(1+\sin\theta \right)^2}\ =&2\sqrt{\left(1-\sin\theta\right)\left(1+\sin\theta \right)^3}\ =&2\sqrt{\dfrac{1}{3}\cdot \left(3-3\sin\theta\right)\left(1+\sin\theta\right)^3}\ \leqslant& 2\sqrt{\dfrac{1}{3}\cdot \left[\dfrac{\left(3-3\sin\theta\right)+\left(1+\sin\theta\right)+\left(1+\sin\theta\right)+\left(1+\sin\theta\right)}{4}\right]^2 }\ =&\dfrac{3\sqrt{3}}{2}. \end{split}$$
因此当$\sin\theta=\dfrac{1}{2}$时,所求面积取得最大值$\dfrac{3\sqrt3}{2}$.
法二 同法一,可得 $$S_{\triangle OAB}=2\cos\theta\left(1+\sin\theta\right).$$ 设上述表达式为$y=f(\theta),\theta\in\left[0,\dfrac{\pi}{2} \right)$,求导可得 $$f'(\theta)=2\left(1-2\sin\theta\right)\left(1+\sin\theta\right),\theta\in\left[0,\dfrac{\pi}{2} \right).$$ 因此当$\theta=\dfrac{\pi}{6}$时,$f(\theta)$也即$S_{\triangle OAB}$取得最大值$\dfrac{3\sqrt{3}}{2}$.