12. The maximum value of the function $y=\sqrt{8 x-x^{2}}-\sqrt{114 x-x^{2}-48}$ is $\qquad$ , and the minimum value is $\qquad$ .
12. $2 \sqrt{3}, 0$.
\[
\left\{\begin{array}{l}
8 x-x^{2} \geqslant 0 \\ 
14 x-x^{2}-48 \geqslant 0,
\end{array}\right.
\]
we get $6 \leqslant x \leqslant 8, f(x)=\sqrt{x(8-x)}-\sqrt{(8-x)(x-6)}=$ 
$\sqrt{8-x}(\sqrt{x}-\sqrt{x-6})=\frac{6 \sqrt{8-x}}{\sqrt{x}+\sqrt{6-x}}, x \in[6,8], f(x)$ is a decreasing function on $[6,8]$, so $f_{\max }=f(6)=2 \sqrt{3}, f_{\min }=f(8)=0$