Example 12. In $\triangle \mathrm{ABC}$, $\angle \mathrm{B}=30^{\circ}, \angle \mathrm{C}=60^{\circ}$, $AC=a$, points $\mathrm{P}$ and $\mathrm{Q}$ start moving from $\mathrm{A}$ along the perimeter of the triangle. The speed of $\mathrm{Q}$ is three times the speed of $\mathrm{P}$.
(1) Let the length of $AP$ be $x$, and the area of $\triangle APQ$ be $y$. Express $y$ in terms of $x$, and draw its graph;
(2) Find the maximum value of $y$ and the corresponding $x$ value.
Given $\mathrm{AC}=\mathrm{a}, \angle \mathrm{B}=30^{\circ}, \angle \mathrm{C}=60^{\circ}$, so!
$$
A B=\sqrt{3} a, B C=2 a \text {. }
$$
(1) When $0 \leqslant x \leqslant \frac{a}{3}$, $\mathrm{P}$ is on $\mathrm{AB}$, $\mathrm{Q}$ is on $\mathrm{AC}$.

So A - A, ACi-3 $\alpha$,
$$
\begin{aligned}
\therefore y^{\prime} & =S, P Q \\
& =\frac{1}{2} x \cdot 3 x=\frac{3}{2} x^{2}
\end{aligned}
$$
$$
\text { For } \frac{1}{3} a \leqslant x \leqslant a \text {, } \mathrm{P} \text { is on: } \mathrm{AB} \text {, } \mathrm{Q} \text { is on CB. }
$$

So $\mathrm{AP}=x, B P=\sqrt{3 a}-x, C Q=3 x-a$, $B Q=2 a-(3 a-a)=3(a-x)$.
$$
\begin{aligned}
\therefore \quad & =S_{\triangle A P Q}=S_{\triangle A B C}-S_{\triangle B P Q}-S_{\triangle A C Q} \\
& =\frac{1}{2} a \cdot \sqrt{3 a}-\frac{1}{2}(\sqrt{3 a}-x) \\
& 3(a-x) \sin 30^{\circ}-\frac{1}{2} a(3 x-a) \sin 60^{\circ} \\
& =3\left(a x-x^{2}\right) \\
& =\frac{3}{16} a^{2}-\frac{3}{4}\left(x-\frac{a}{2}\right)^{2}
\end{aligned}
$$
$$
\begin{array}{l}
x+3 x=a+2 a+\sqrt{3} a \\
\therefore x=3+3^{3} .
\end{array}
$$

Therefore, when $a \leqslant x \leqslant 3+\sqrt{3}$ ais,
$$
\mathbf{y}=0 \text {. }
$$

From (1)(2)(3) we get the above.
(2) From the graph, when $x=\frac{a}{2}$,
$$
y_{\max }=\frac{3}{16} \mathrm{a}^{2} \text {. }
$$