Example 8 As shown in the figure, given $A(0, a), B(0, b), (0<a<b)$ are fixed points, find a point $C$ on the positive half-axis of the $x$-axis, such that $\angle ACB$ is maximized, and find the maximum value.
Analyze: Using the dot product, establish a functional relationship about point $C$, and then find the maximum or minimum value of the function.

Solution: Let point $C(x, 0), x>0$, then $\overrightarrow{C A}=(0, a)-(x, 0)=(-x, a), \overrightarrow{C B}=(0, b)-(x, 0)=(-x, b)$.
So $\overrightarrow{C A} \cdot \overrightarrow{C B}=x^{2}+a b$
$$
|\overrightarrow{C A}|=\sqrt{x^{2}+a^{2}}, |\overrightarrow{C B}|=\sqrt{x^{2}+b^{2}}
$$

Therefore, $\cos \angle A C B=\frac{\overrightarrow{C A} \cdot \overrightarrow{C B}}{|\overrightarrow{C A}||\overrightarrow{C B}|}=\frac{x^{2}+a b}{\sqrt{x^{2}+a^{2}} \cdot \sqrt{x^{2}+b^{2}}}$
Let $t=x^{2}+a b$, then $x^{2}=t-a b$
Thus, $\cos \angle A C B=\frac{t}{\sqrt{\left(t-a b+a^{2}\right)\left(t-a b+b^{2}\right)}}$
$$
\begin{array}{l}
=\frac{t}{\sqrt{t^{2}+(a-b)^{2} t-a b(a-b)^{2}}} \\
=\frac{1}{\sqrt{-a b(a-b)^{2} \cdot \frac{1}{t^{2}}+(a-b)^{2} \cdot \frac{1}{t}+1}}
\end{array}
$$

When $\frac{1}{t}=\frac{1}{2 a b}$, i.e., $t=2 a b$, $\cos \angle A C B$ achieves its minimum value $\frac{2 \sqrt{a b}}{a+b}$. That is, when $x=\sqrt{a b}$, the maximum value of $\angle A C B$ is $\arccos \frac{2 \sqrt{a b}}{a+b}$.