11. 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 $\qquad$ .
$$
\begin{array}{l}
x w+2 y w+3 x y+3 z w+4 x z+5 y z \\
=x y+x z+x w+2(y x+y z+y w)+3(z x+z y+z w)=x(1-x)+2 y(1-y)+3 z(1-z) \\
\leqslant 6 \cdot \frac{1}{4}=\frac{3}{2}, \text{ equality holds when } x=y=z=\frac{1}{2}, w=-\frac{1}{2}.
\end{array}
$$