Example 5 Given a regular hexagon $A B C D E F$ with side length $1, Q R$ is any line segment inside the hexagon parallel to $A B$. Find the maximum area of $\triangle P Q R$ inscribed in the regular hexagon $A B C D E F$ with $Q R$ as the base.
To obtain the maximum area of $\triangle PQR$, point $P$ should be on $DE$, and points $Q, R$ should be on $AF, BC$ respectively. Draw $PH \perp QR$ at point $H$, intersecting $AB$ at point $G$. Draw $AM \perp QR$ and $BN \perp QR$ at points $M, N$ respectively. Let $PH = x$. Then
$$
\begin{array}{l}
H G = \sqrt{3} - x, \\
M Q = N R = \frac{\sqrt{3}}{3} A M = \frac{\sqrt{3}}{3}(\sqrt{3} - x) = 1 - \frac{\sqrt{3}}{3} x, \\
Q R = 2\left(1 - \frac{\sqrt{3}}{3} x\right) + 1 = 3 - \frac{2 \sqrt{3}}{3} x. \\
\text{Therefore, } S_{\triangle P Q R} = \frac{1}{2}\left(3 - \frac{2 \sqrt{3}}{3} x\right) x \\
= -\frac{\sqrt{3}}{3}\left(x - \frac{3 \sqrt{3}}{4}\right)^{2} + \frac{9 \sqrt{3}}{16}.
\end{array}
$$

When $x = \frac{3 \sqrt{3}}{4}$, i.e., when $Q, R$ are the midpoints of $AF, BC$ respectively, $S_{\triangle P Q R}$ reaches its maximum value $\frac{9 \sqrt{3}}{16}$.