Question

Suppose $x_1,x_2,\dots ,x_{49}$ are real numbers such that $x_1^2+2x_2^2+\cdots +49x_{49}^2=1.$ The maximum value of $x_1+2x_2+\cdots +49x_{49}$ is
(correct answer + 2, wrong answer 0)

Answer 1

$(x_1+2x_2+\cdots +49x_{49}{)}^2$
$=(1\cdot x_1+\sqrt{2}\cdot \sqrt{2}{x}_2+\cdots +\sqrt{49}\cdot \sqrt{49}{x}_{49}{)}^2$
$\le (1+2+\cdots +49)(x_1^2+2x_2^2+\cdots +49x_{49}^2)$
$=\frac{49\times 50}{2}\times 1={35}^2$
The equality holds if and only if $x_1=\cdots =x_{49}=\pm \frac{1}{35}$