$13 \cdot 36$ For real number $x$, the function
$$
f(x)=\sqrt{8 x-x^{2}}-\sqrt{14 x-x^{2}-48}
$$

has the maximum value of
(A) $\sqrt{7}-1$.
(B) 3 .
(C) $2 \sqrt{3}$.
(D) 4 .
(E) $\sqrt{55}-\sqrt{5}$.
(44th American High School Mathematics Examination, 1993)
[Solution] Since $f(x)=\sqrt{(8-x) x}-\sqrt{(8-x)(x-6)}$, it is a real number if and only if $6 \leqslant x \leqslant 8$. Also,
$$
\begin{aligned}
f(x) & =\sqrt{8-x}(\sqrt{x}-\sqrt{x-6}) \\
& =\sqrt{8-x} \cdot \frac{(\sqrt{x}-\sqrt{x-6})(\sqrt{x}+\sqrt{x-6})}{\sqrt{x}+\sqrt{x-6}} \\
& =\frac{6 \sqrt{8-x}}{\sqrt{x}+\sqrt{x-6}}
\end{aligned}
$$

holds for all $6 \leqslant x \leqslant 8$.
When $x=6$, the numerator of the last expression is maximized, and the denominator is minimized.
Therefore, the value of the fraction is maximized, which is $f(6)=\frac{6 \sqrt{2}}{\sqrt{6}}=2 \sqrt{3}$.
Hence, the answer is $(C)$.