2. In the Cartesian coordinate system, the coordinates of the three vertices of the right triangle $Rt \triangle OAB$ are $O(0,0), A(1,0)$, and $B(1,2)$. Take any point $C(x, 2x) (0<x<1)$ on the hypotenuse, and draw $CD \perp OA$ at $D$, and $CE \perp AB$ at $E$. Let the area of $\triangle OCD$ be $S_{1}(x)$, the area of rectangle $CDAE$ be $S_{2}(x)$, and the area of $\triangle CEB$ be $S_{3}(x)$ (as shown in Figure 5). For the same $x$, let $f(x)$ be the maximum value of $S_{1}(x)$, $S_{2}(x)$, and $S_{3}(x)$. When $C$ varies on the line segment $OB$, the minimum value of $f(x)$ is $\qquad$
$2 \cdot \frac{1}{9}$.
From the area formula, it is easy to get
$$
\begin{array}{l}
S_{1}(x)=\frac{1}{2} O D \cdot D C \\
=x^{2}(0<x<1), \\
S_{2}(x)=D C \cdot D A \\
=2 x(1-x) \\
(0<x<1), \\
S_{3}(x)=\frac{1}{2} C E \cdot E B \\
=(1-x)^{2}(0<x<1) .
\end{array}
$$

With the help of the figure (Figure 16), it is convenient to derive
$$
f(x)=\left\{\begin{array}{lc}
(1-x)^{2}, & 0<x \leqslant \frac{1}{3} \\
2 x(1-x), & \frac{1}{3}<x<\frac{2}{3} \\
x^{2}, & \frac{2}{3} \leqslant x<1
\end{array}\right.
$$

It is evident that when $x=\frac{1}{3}$ or $\frac{2}{3}$, $f(x)$ has the minimum value $\frac{4}{9}$.