18 (14 points) A horizontal L-shaped corridor is shown in the figure inside a building. The width of both corridors is 3 meters. A piece of equipment with a rectangular cross-section needs to be moved horizontally into the L-shaped corridor. If the width of the rectangular cross-section of the equipment is 1 meter and the length is 7 meters, can the equipment be moved horizontally into the L-shaped corridor?
18 Establish a Cartesian coordinate system with the lines $O B$ and $O A$ as the $x$-axis and $y$-axis, respectively. The problem then transforms into: first, finding the minimum length of the segment $A B$ intercepted by the positive $x$-axis and the positive $y$-axis, where $A B$ is a tangent to the circle with center $M(3,3)$ and radius 1; then, comparing this minimum length with 7.

Let the equation of the line $A B$ be $\frac{x}{a}+\frac{y}{b}=1$. Since this line is tangent to the circle $(x-3)^{2}+(y-3)^{2}=1$, we have
$$
\left|\frac{3}{a}+\frac{3}{b}-1\right|=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}.
$$
1 Since the origin $O(0,0)$ and the point $M(3,3)$ are on opposite sides of the line $\frac{x}{a}+\frac{y}{b}=1$, we have $\frac{3}{a}+\frac{3}{b}-1>0$, so (1) can be rewritten as
$$
\sqrt{a^{2}+b^{2}}=3(a+b)-a b.
$$

Next, we need to find the minimum value of $|A B|=\sqrt{a^{2}+b^{2}}$ for $a>0, b>0$.
Let $a=r \sin \theta, b=r \cos \theta, r>0, \theta \in\left(0, \frac{\pi}{2}\right)$, substituting into (2) gives
$$
r=\frac{3(\sin \theta+\cos \theta)-1}{\sin \theta \cos \theta}.
$$

Let $t=\sin \theta+\cos \theta$, since $\theta \in\left(0, \frac{\pi}{2}\right)$, we have $t \in(1, \sqrt{2}]$, substituting into (3) gives $r=\frac{6 t-2}{t^{2}-1}$, i.e., $r t^{2}-6 t-r+2=0$. Let $f(t)=r t^{2}-6 t-r+2, t \in(1, \sqrt{2}], r>0$.

Since $f(1)=-4<0$, the equation $r t^{2}-6 t-r+2=0$ has a solution in $t \in(1, \sqrt{2}]$, i.e., $f(\sqrt{2})=r-6 \sqrt{2}+2 \geqslant 0$, solving for $r$ gives $r \geqslant 6 \sqrt{2}-2$, which holds when and only when $t=\sqrt{2}$, i.e., $\theta=\frac{\pi}{4}$.

This indicates that the maximum length of a rectangle with a width of 1 meter that can pass through the corridor is $r_{\text {min }}=6 \sqrt{2}-2<6 \times \frac{3}{2}-2=7$ meters, hence the equipment cannot be moved horizontally through the corridor.