Question

tìm giá trị lớn nhất của \(\left|x+2y+3z\right|\)

với (x^2+y^2+z^2=1)

Answer 1

\(A=\left|x+2y+3z\right|\Rightarrow A^2\le\left(1+2^2+3^2\right)\left(x^2+y^2+z^2\right)=14\Rightarrow A\le\sqrt{14}\)

\(max_A=\sqrt{14}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y}{2}=\dfrac{z}{3}\\x^2+y^2+z^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x;y;z\right)=\left(\dfrac{1}{\sqrt{14}};\sqrt{\dfrac{2}{7}};\dfrac{3}{\sqrt{14}}\right)\\\left(x;y;z\right)=\left(-\dfrac{1}{\sqrt{14}};-\sqrt{\dfrac{2}{7}};-\dfrac{3}{\sqrt{14}}\right)\end{matrix}\right.\)