每日一题[967]逐步消元

已知函数$f(x)=x^2+ax+b$($a,b\in\mathbb R$)在区间$(0,1]$上有零点\(x_0\)，则$$ab\left(\dfrac{x_0}4+\dfrac{1}{9x_0}-\dfrac 13\right)$$的最大值是________．

分析与解　正确答案是$\dfrac{1}{144}$．

因为$x_0^2+ax_0+b=0$，所以$$b=-x_0^2-ax_0.$$根据题意，有\[\begin{split}ab\left(\dfrac{x_0}4+\dfrac{1}{9x_0}-\dfrac 13\right)&=a\cdot \left(-x_0^2-ax_0\right)\cdot \left(\dfrac{x_0}4+\dfrac{1}{9x_0}-\dfrac 13\right)\\&=\dfrac {1}{36}a\left(-x_0-a\right)\left(3x_0-2\right)^2\\&\leqslant \dfrac{1}{36}\cdot \dfrac{x_0^2}4\cdot \left(3x_0-2\right)^2\\&=\dfrac{1}{144}\cdot \left[x_0\left(3x_0-2\right)\right]^2\\&\leqslant \dfrac{1}{144},\end{split}\]等号当$x_0=1$，$a=-\dfrac 12$时取得．因此所求的最大值为$\dfrac{1}{144}$．