11. Let real numbers $x, y, z, w$ satisfy $x+y+z+w=1$. Then the maximum value of $M=x w+2 y w+3 x y+3 z w+4 x z+5 y z$ is $\qquad$ .
11. $\frac{3}{2}$.

It is easy to know that for any real number $t$, we have $t(1-t) \leqslant \frac{1}{4}$, with equality holding if and only if $t=\frac{1}{2}$.
Notice that,
$$
\begin{aligned}
M= & x(y+z+w)+2 y(x+z+w)+ \\
& 3 z(x+y+w) \\
= & x(1-x)+2 y(1-y)+3 z(1-z) \\
\leqslant & \frac{1}{4}+2 \times \frac{1}{4}+3 \times \frac{1}{4}=\frac{3}{2} .
\end{aligned}
$$

Equality holds if and only if $x=y=z=\frac{1}{2}, w=-\frac{1}{2}$.

Therefore, the maximum value of $M$ is $\frac{3}{2}$.