2. Given non-negative real numbers $x, y, z$ satisfy $x+y+z=1$. Then the maximum value of $t=2xy+yz+2zx$ is ( ).
(A) $\frac{4}{7}$
(B) $\frac{5}{9}$
(C) $\frac{9}{16}$
(D) $\frac{12}{25}$
2. A.

Notice that,
$$
\begin{aligned}
t & =2 x y+y z+2 z x=2 x(y+z)+y z \\
& \leqslant 2 x(y+z)+\frac{1}{4}(y+z)^{2} \\
& =2 x(1-x)+\frac{1}{4}(1-x)^{2} \\
& =-\frac{7}{4}\left(x-\frac{3}{7}\right)^{2}+\frac{4}{7} .
\end{aligned}
$$

When $x=\frac{3}{7}, y=z=\frac{2}{7}$, the equality holds.
At this point, $t=2 x y+y z+2 z x$ achieves its maximum value $\frac{4}{7}$.