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Find the maximum of the the function $$f(\bar x)=\sqrt{x_1^2+2x_2^2+......2016x_{2016}^2}$$ $\bar x=(x_1,x_2,....x_{2016}) \in R^{2016}$
where the domain of $f$ is $$\bar x=(x_1,x_2,....x_{2016}) \in R^{2016} :\sum_{n=1}^{2016}x_n^2=1$$.
Sol:
$f^2(\bar x)={x_1^2+2x_2^2+......2016x_{2016}^2}=...