求函數 \(2x+\frac{27}{x^2}\),\((x>0)\)的最小值
解﹕\(\because x>0\)
\(\therefore y=2x+\frac{27}{x^2}\)
\(=x+x+\frac{27}{x^2}\geq 3\sqrt[3]{x\cdot x\cdot \frac{27}{x^2}}=9\)
當且僅當 \(x=\frac{27}{x^2}\),即\(x=3\)時,\(y\)的最小值是\(9\)
分類
求函數 \(2x+\frac{27}{x^2}\),\((x>0)\)的最小值
解﹕\(\because x>0\)
\(\therefore y=2x+\frac{27}{x^2}\)
\(=x+x+\frac{27}{x^2}\geq 3\sqrt[3]{x\cdot x\cdot \frac{27}{x^2}}=9\)
當且僅當 \(x=\frac{27}{x^2}\),即\(x=3\)時,\(y\)的最小值是\(9\)