求函數 \(y=x^2(12-5x)\),\((0<x<\frac{12}{5})\) 的最大值
解﹕\(\because 0<x<\frac{12}{5}, \therefore 12-5x>0 \)
\(\therefore y=x^2(12-5)\)
\(=\frac{4}{25}\cdot\frac{5x}{2}\cdot\frac{5x}{2}\cdot (12-5x)\leq \frac{4}{25}(\frac{\frac{5x}{2}+\frac{5x}{2}+12-5x}{3})^3=\frac{256}{25}\)
當且僅當\(\frac{5x}{2}=12-5x\),即 \(x=\frac{8}{5}\)時，\(y\)有最大值是\(\frac{256}{25}\)
#P13