设函数 $f(x,y)=-6xy+\dfrac{7}{2}(x+y)-2$，则 $\min\limits_{x\in[0,1]}\left\{\max\limits_{y\in[0,1]}\left\{f(x,y)\right\}\right\}=$ \((\qquad)\)
【难度】
【出处】
2016年清华大学自主招生暨领军计划试题
【标注】
【答案】
BD
【解析】
由 $x,y$ 的对称性可知，$\min\limits_{x\in[0,1]}\left\{\max\limits_{y\in[0,1]}\left\{f(x,y)\right\}\right\}=\min\limits_{y\in[0,1]}\left\{\max\limits_{x\in[0,1]}\left\{f(x,y)\right\}\right\}$．而\begin{align*}
\min\limits_{x\in[0,1]}\left\{\max\limits_{y\in[0,1]}\left\{f(x,y)\right\}\right\}
&=\min\limits_{x\in[0,1]}\left\{\max\limits_{y\in[0,1]}\left\{\left(\dfrac{7}{2}-6x\right)y+\left(\dfrac{7}{2}x-2\right)\right\}\right\}\\
&=\min\limits_{x\in[0,1]}\left\{\max\left\{\left(\dfrac{7}{2}-6x\right)\cdot 0+\left(\dfrac{7}{2}x-2\right), \left(\dfrac{7}{2}-6x\right)\cdot 1+\left(\dfrac{7}{2}x-2\right)\right\}\right\}\\
&=\min\limits_{x\in[0,1]}\left\{\max\left\{\dfrac{7}{2}x-2, -\dfrac{5}{2}x+\dfrac{3}{2}\right\}\right\}\\
&=\dfrac{1}{24},
\end{align*}当且仅当 $x=\dfrac{7}{12}$ 或 $y=\dfrac{7}{12}$ 时等号成立．
\min\limits_{x\in[0,1]}\left\{\max\limits_{y\in[0,1]}\left\{f(x,y)\right\}\right\}
&=\min\limits_{x\in[0,1]}\left\{\max\limits_{y\in[0,1]}\left\{\left(\dfrac{7}{2}-6x\right)y+\left(\dfrac{7}{2}x-2\right)\right\}\right\}\\
&=\min\limits_{x\in[0,1]}\left\{\max\left\{\left(\dfrac{7}{2}-6x\right)\cdot 0+\left(\dfrac{7}{2}x-2\right), \left(\dfrac{7}{2}-6x\right)\cdot 1+\left(\dfrac{7}{2}x-2\right)\right\}\right\}\\
&=\min\limits_{x\in[0,1]}\left\{\max\left\{\dfrac{7}{2}x-2, -\dfrac{5}{2}x+\dfrac{3}{2}\right\}\right\}\\
&=\dfrac{1}{24},
\end{align*}当且仅当 $x=\dfrac{7}{12}$ 或 $y=\dfrac{7}{12}$ 时等号成立．
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