11. The maximum value of the function $y=2 x+\sqrt{1-2 x}$ is
11. $\frac{5}{4}$.

Let $t=\sqrt{1-2 x}, t \geqslant 0$, then $x=\frac{1}{2}\left(1-t^{2}\right)$, so $y=1-t^{2}+t=-\left(t-\frac{1}{2}\right)^{2}+\frac{5}{4}$.