MT321分类线性规划

Posted 2021-11-12 mathstudy

若二次函数$f(x)=ax^2+bx+c(a,b,c>0)$有零点,则$\\min\\{\\dfrac{b+c}{a},\\dfrac{c+a}{b},\\dfrac{a+b}{c}\\}$ 的最大值为____

由题意$b^2\\ge 4ac,$由$a,c$的对称性只需考虑$b=max\\{a,b,c\\}\\vee c=\\max\\{a,b,c\\}$.
当$b=max\\{a,b,c\\}$时$\\min\\{\\dfrac{b+c}{a},\\dfrac{c+a}{b},\\dfrac{a+b}{c}\\}=\\dfrac{c+a}{a}$
设$\\dfrac{a}{b}=x,\\dfrac{c}{b}=y$故
\\begin{equation}
\\left\\{ \\begin{aligned}
0<x& \\le1\\\\
0<y&\\le1\\\\
0<y&\\le\\dfrac{1}{4x}\\\\
\\end{aligned} \\right.
\\end{equation}
由线性规划可知$z=x+y$在$(1,\\dfrac{1}{4})$和$(\\dfrac{1}{4},1)$处同时取最大值$\\dfrac{5}{4}$.
当$c=\\max\\{a,b,c\\}$时$\\min\\{\\dfrac{b+c}{a},\\dfrac{c+a}{b},\\dfrac{a+b}{c}\\}=\\dfrac{a+b}{c}$
设$\\dfrac{b}{a}=x,\\dfrac{c}{a}=y$故
\\begin{equation}
\\left\\{ \\begin{aligned}
0<x& \\le1\\\\
0<y&\\le1\\\\
y&\\le x\\\\
0<y&\\le\\dfrac{1}{4x}\\\\
\\end{aligned} \\right.
\\end{equation}
由线性规划可知$z=x+y$在$(1,\\dfrac{1}{4})$时取最大值$\\dfrac{5}{4}$.