求最大值(練1)

已知 $0<x<\frac{2}{5}$，求下列式子的最大值.

(1) $x(2-5x)$

解﹕$\because 0<x<\frac{2}{5}$

$x(2-5x)=\frac{1}{5}\cdot 5x(2-5x)\leq\frac{1}{5}(\frac{5x+(2-5x)}{2})^2=\frac{1}{5}$

當且僅當 $5x=2-5x$，即 $x=\frac{1}{5}$ 時，$x(2-5x)$的最大值是$\frac{1}{5}$

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(2) $x^2(2-5x)$

解﹕$\because 0<x<\frac{2}{5}$

$x^2(2-5x)=\frac{4}{25}\cdot\frac{5x}{2}\cdot\frac{5x}{2}(2-5x)\leq\frac{4}{25}\left(\frac{\frac{5x}{2}+\frac{5x}{2}+(2-5x)}{3}\right)^3$

$=\frac{4}{25}\cdot\frac{8}{27}=\frac{32}{675}$

當且僅當 $\frac{5x}{2}=2-5x$，即 $x=\frac{4}{15}$ 時，$x^2(2-5x)$的最大值是$\frac{32}{675}$

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(3) $x(2-5x)^2$

解﹕$\because 0<x<\frac{2}{5}$

$x(2-5x)^2$

$=\frac{1}{10}\cdot 10x(2-5x)(2-5x)$

$\leq\frac{1}{10}\left(\frac{10+(2-5x)+(2-5x)}{3}\right)^3$

$\frac{1}{10}\cdot(\frac{14}{3})^3=\frac{1}{10}\cdot\frac{2744}{27}=\frac{1372}{135}$

當且僅當 $10x=2-5x$，即$x=\frac{2}{15}$時，$x(2-5x)^2$的最大值是$\frac{1372}{135}$