11. (20 points) Given non-negative real numbers $x, y, z$ satisfying $x+y+z=1$. Find the maximum and minimum values of $x^{3}+2 y^{2}+\frac{10}{3} z$.
11. The maximum value is $\frac{10}{3}$, the minimum value is $\frac{14}{27}$.

Given $x, y, z \in [0,1]$, we know $x^{3} \leqslant x, y^{2} \leqslant y$.
Thus, $x^{3} + 2 y^{2} + \frac{10}{3} z \leqslant x + 2 y + \frac{10}{3} z$
$$
\leqslant \frac{10}{3}(x + y + z) = \frac{10}{3}.
$$

When $x = y = 0, z = 1$, the above expression achieves the maximum value $\frac{10}{3}$.
Also, $x^{3} + \frac{8}{27} + \frac{8}{27} \geqslant 3 \cdot x \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{3} x$,
$2 y^{2} + \frac{2}{9} \geqslant \frac{4}{3} y$,
then $\left(x^{3} + \frac{16}{27}\right) + \left(2 y^{2} + \frac{2}{9}\right) + \frac{10}{3} z$
$\geqslant \frac{4}{3} x + \frac{4}{3} y + \frac{10}{3} z \geqslant \frac{4}{3}(x + y + z) = \frac{4}{3}$
$\Rightarrow x^{3} + 2 y^{2} + \frac{10}{3} z \geqslant \frac{14}{27}$.
When $x = \frac{2}{3}, y = \frac{1}{3}, z = 0$, the above expression achieves the minimum value $\frac{14}{27}$.