$17 \cdot 171$ In $\triangle A B C$, $A B+A C=10$, and $\angle A=30^{\circ}$, then the area of this triangle
(A) has both a maximum and a minimum value.
(B) has a minimum value, but no maximum value.
(C) has a maximum value, but no minimum value.
(D) has neither a maximum nor a minimum value.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Let $AB = x$, then $AC = 10 - x$,
By $S_{\triangle ABC} = \frac{1}{2} x(10 - x) \sin 30^{\circ} = -\frac{1}{4} x^2 + \frac{5}{2} x$
$$
= -\frac{1}{4}(x - 5)^2 + \frac{25}{4},
$$

Since $0 < x < 10$, and $5 \in (0, 10)$,
then
$$
\begin{array}{l}
S_{\max} = \left.S\right|_{x=5} = \frac{25}{4}. \\
S_{\triangle ABC} \leqslant S_{\max} = \frac{25}{4}.
\end{array}
$$

As $x \rightarrow 10$, $S_{\triangle ABC} \rightarrow 0$, thus $S_{\text{min}}$ does not exist. Therefore, the answer is $(C)$.