5. The maximum value of the function $y=2 x-5+\sqrt{11-3 x}$ is $\qquad$ .
5. $\frac{65}{24}$.

Let $\sqrt{11-3 x}=t$. Then
$$
\begin{array}{l}
6 y=12 x-30+6 \sqrt{11-3 x} \\
=-4 t^{2}+6 t+14 \\
=-\left(2 t-\frac{3}{2}\right)^{2}+\frac{65}{4} \leqslant \frac{65}{4} .
\end{array}
$$

Thus, $y \leqslant \frac{65}{24}$.
When $t=\frac{3}{4}$, i.e., $x=\frac{167}{48}$, the equality holds.