(5) The maximum value of the function $y=2 x-5+\sqrt{11-3 x}$ is $\qquad$ .
(5) $\frac{65}{24}$ Hint: Let $\sqrt{11-3 x}=t$, then
$$
\begin{aligned}
6 y & =12 x-30+6 \sqrt{11-3 x} \\
& =-4(11-3 x)+6 \sqrt{11-3 x}+14 \\
& =-4 t^{2}+6 t+14 \\
& =-\left(2 t-\frac{3}{2}\right)^{2}+\frac{65}{4} \leqslant \frac{65}{4} .
\end{aligned}
$$

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