MT【179】最大最小老问题
求$\max\{x^2+2y+20,y^2-6x+12\}$的最小值______
提示:$4t\ge 3(x^2+2y+20)+y^2-6x+12=3(x-1)^2+(y+3)^2+60\ge 60,t\ge 15$
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求$\max\{x^2+2y+20,y^2-6x+12\}$的最小值______
提示:$4t\ge 3(x^2+2y+20)+y^2-6x+12=3(x-1)^2+(y+3)^2+60\ge 60,t\ge 15$