Question

Tìm GTLN của \(y=\left|x\right|\sqrt{9-x^2}\)

Answer 1

đk <=>\(\left\{{}\begin{matrix}\left|x\right|\le3\\y\ge0\end{matrix}\right.\)

\(\left|x\right|.\sqrt{9-x^2}\le\dfrac{x^2+9-x^2}{2}=\dfrac{9}{2}\)

đẳng thức khi

\(x^2=9-x^2=>x=\pm\dfrac{3\sqrt{2}}{2}\)