Example 7. In an equilateral $\triangle ABC$ with side length $a$, moving points $P$, $Q$, $R$ start simultaneously from $A$, $B$, $C$ and move along $AB$, $BC$, $CA$ in a uniform linear motion in the counterclockwise direction. It is known that point $P$ takes 1 second to travel from $A$ to $B$, point $Q$ takes 2 seconds to travel from $B$ to $C$, and point $R$ takes 3 seconds to travel from $C$ to $A$. Within one second, after how much time does the area of $\triangle PQR$ become the smallest? What is the minimum area? (1983, Chongqing Junior High School Competition)
Let the moving point start to move for $t$ seconds, and $S \triangle \mathrm{P} Q \mathrm{R}$ is minimized, then $V_{\mathrm{P}}=a, V_{Q}=\frac{1}{2} a, V_{\mathrm{B}}=\frac{1}{3} a, A P=a t$,
$$
\begin{array}{l}
P B=a(1-t), B Q=\frac{a t}{2}, Q C=a\left(1-\frac{t}{2}\right), \\
C R=\frac{a t}{3}, R A=a\left(1-\frac{t}{3}\right) .
\end{array}
$$

Thus, $S_{\triangle \triangle P R}+S \triangle \mathrm{BQP}+S \triangle \mathrm{CRQ}$
$$
\begin{aligned}
& =\frac{\sqrt{3}}{4} a^{2}\left(-t^{2}+\frac{11}{6} t\right) \\
& =\frac{\sqrt{3}}{4} a^{2}\left[-\left(t-\frac{11}{12}\right)^{2}+\frac{121}{14 t}\right] . \\
\text { When } t & =\frac{11}{12} \text {, } S \triangle \mathrm{APR}+S \triangle \mathrm{B} 2 \mathrm{P}+S \triangle \mathrm{CR} 2
\end{aligned}
$$

has a maximum value of $\frac{121}{144} \cdot \frac{\sqrt{3}}{4} a^{2}$.
Since $S_{\triangle \mathrm{ABC}}=\frac{1}{2} a^{2} \sin 60^{\circ}=\frac{\sqrt{3}}{4} a^{2}$,

the minimum value of $S \mathrm{PQR}$ is
$$
\frac{\sqrt{3}}{4} a^{2}-\frac{121}{144} \cdot \frac{\sqrt{3}}{4} a^{2}=\frac{23 \sqrt{3}}{576} a^{2} .
$$