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\(\frac{1}{5}\)
(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\(\frac{32}{675}\)
(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\(\frac{1372}{135}\)