Example 11 A right-angled corridor is 1.5 meters wide, as shown in Figure 1-2. There is a flexible-handling cart with a rectangular platform that is 1 meter wide. What is the maximum length of the platform that allows the cart to smoothly pass through the right-angled corridor?
The problem scenario and question are concise and clear, and are often encountered in daily life. How to choose effective methods and means, through observation, analysis, abstraction, and modeling, to apply the relevant knowledge learned to solve problems. This is a good example.

Analysis: Intuitively, it is believed that when the flatbed truck forms a $45^{\circ}$ angle with the wall, the truck can smoothly pass the corner, and the length of the truck body (as shown in Figure 1-2) $C D=A B-2 B D=3 \sqrt{2}-2$ (meters). Why?

Consider a horizontally placed bamboo pole passing through point $M$, with the two endpoints $A, B$ on the wall (as shown in Figure 1-3). If the angle $\theta$ between the bamboo pole and the wall is greater than or less than $45^{\circ}$, it cannot smoothly pass the corner. Only when $\theta$ equals $45^{\circ}$ can it pass, at which point the length of the bamboo pole $A B$ is the shortest. If $\theta \neq 45^{\circ}$, the length of $A B$ exceeds a certain value, and the bamboo pole can only pass in one direction.

Based on the above analysis, the problem can be abstracted as follows:
In a Cartesian coordinate system, a line $l$ passing through the fixed point $M\left(\frac{3}{2}, \frac{3}{2}\right)$ intersects the positive x-axis and y-axis at points $A, B$ respectively. Try to find the minimum length of the line segment $A B$ (as shown in Figure 1-4).
When the length of $A B$ is the smallest, the length of the flatbed truck $C D$ is the maximum length that can smoothly pass the corner.

To solve the above extremum problem, the following ideas can be used:
(1) Let the line $l$ intersect the x-axis and y-axis at points $A(a, 0), B(0, b)$ respectively $(a>0, b>0)$, and it passes through the point $M\left(\frac{3}{2}, \frac{3}{2}\right)$, i.e., it satisfies the condition: $\frac{3}{2 a}+\frac{3}{2 b}=1$, which is $3(a+b)=2 a b$. Also, $A B=\sqrt{a^{2}+b^{2}}$. Thus, the problem is transformed into: when $a>0, b>0$ and $3(a+b)=2 a b$, find the minimum value of $A B=\sqrt{a^{2}+b^{2}}$.
(2) Let $\angle O A B=\theta, \theta \in\left(0, \frac{\pi}{2}\right)$, then $A B=A M+M B=\frac{3}{2}\left(\frac{1}{\sin \theta}+\frac{1}{\cos \theta}\right)$. The problem is transformed into: when $\theta \in\left(0, \frac{\pi}{2}\right)$, find the minimum value of the function $g(\theta)=\frac{3}{2}\left(\frac{1}{\sin \theta}+\frac{1}{\cos \theta}\right)$.
Solve $A B=g(\theta)=\frac{3}{2}\left(\frac{1}{\sin \theta}+\frac{1}{\cos \theta}\right)=\frac{3}{2} \sqrt{\frac{1+\sin 2 \theta}{\frac{1}{4} \sin ^{2} 2 \theta}}=3 \sqrt{\frac{1}{\sin ^{2} 2 \theta}+\frac{1}{\sin 2 \theta}}$.
Since $\theta \in\left(0, \frac{\pi}{2}\right)$, we have $0<\sin 2 \theta \leqslant 1$. It is easy to see that when $\sin 2 \theta=1$, $g(\theta)$ has the minimum value $3 \sqrt{2}$. That is, $A B$ has the minimum value $3 \sqrt{2}$, at which point $\theta=\frac{\pi}{4}$.