6. Let $\frac{1}{3} \leqslant x \leqslant 5$. Then the maximum value of $3 \sqrt{5-x}+2 \sqrt{3 x-1}$ is ( ).
(A) $2 \sqrt{14}$
(B) $\frac{5 \sqrt{14}}{2}$
(C) $5 \sqrt{2}$
(D) $7 \sqrt{2}$
6.D.

Let $\sqrt{5-x}=a, \sqrt{3 x-1}=b$. Squaring both equations and eliminating $x$ yields
$$
3 a^{2}+b^{2}=14 \text {. }
$$

Now let $y=3 \sqrt{5-x}+2 \sqrt{3 x-1}$, then
$$
y=3 a+2 b \text {. }
$$

From equation (2), we get $a=\frac{y-2 b}{3}$, substituting this into equation (1) and simplifying yields
$$
7 b^{2}-4 y b+\left(y^{2}-42\right)=0 \text {. }
$$

Considering the above equation as a quadratic equation in $b$, since $b$ is a real number, we have
$$
\Delta=16 y^{2}-28\left(y^{2}-42\right)=-12 y^{2}+1176 \geqslant 0 \text {. }
$$

Solving this, we get $y \leqslant 7 \sqrt{2}$.
$$
\text { When } y=7 \sqrt{2} \text {, } b=2 \sqrt{2}, x=3, a=\sqrt{2} \text {. }
$$