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,
&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\(\dfrac{3\sqrt{3}}{2}\).
