18. Suppose $x_{1}, x_{2}, \ldots, x_{49}$ are real numbers such that
$$
x_{1}^{2}+2 x_{2}^{2}+\cdots+49 x_{49}^{2}=1 .
$$

Find the maximum value of $x_{1}+2 x_{2}+\cdots+49 x_{49}$.
18. Answer: 35 .
$$
\begin{aligned}
\left(x_{1}+2 x_{2}+\cdots+49 x_{49}\right)^{2} & =\left(1-x_{1}+\sqrt{2} \cdot \sqrt{2} x_{2}+\cdots+\sqrt{49} \cdot \sqrt{49} x_{49}\right)^{2} \\
& \leq(1+2+\cdots+49)\left(x_{1}^{2}+2 x_{2}^{2}+\cdots+49 x_{49}^{2}\right) \\
& =\frac{49 \times 50}{2} \times 1=35^{2} .
\end{aligned}
$$

The equality holds if and only if $x_{1}=\cdots=x_{49}= \pm \frac{1}{35}$.