18. (14 points) A horizontal L-shaped corridor in a building is shown in Figure 2. The width of both corridors is $3 \mathrm{~m}$, and 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 $1 \mathrm{~m}$ and the length is $7 \mathrm{~m}$, can the equipment be moved horizontally into the L-shaped corridor?
18. Establish a rectangular 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, which is a tangent to the circle with center $M(3,3)$ and radius 1, and then comparing this minimum value with 7. Let
$$
l_{A B}: \frac{x}{a}+\frac{y}{b}=1 .
$$

Since this line is tangent to the circle $(x-3)^{2}+(y-3)^{2}=1$, we have
$$
\left|\frac{3}{a}+\frac{3}{b}-1\right|=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}} .
$$

Since the origin $O(0,0)$ and $M(3,3)$ are on opposite sides of the line $\frac{x}{a}+\frac{y}{b}=1$, it follows that $\frac{3}{a}+\frac{3}{b}-1>0$.
Thus, equation (1) can be rewritten as
$$
\sqrt{a^{2}+b^{2}}=3(a+b)-a b \text {. }
$$

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

Let $t=\sin \theta+\cos \theta$.
Since $\theta \in\left[0, \frac{\pi}{2}\right]$, we have $t \in(1, \sqrt{2}]$.
Substituting into equation (3) gives
$$
r=\frac{6 t-2}{t^{2}-1}=\frac{4}{t+1}+\frac{2}{t-1} \geqslant 6 \sqrt{2}-2 \text {. }
$$

The equality holds if and only if $t=\sqrt{2}$, i.e., $\theta=\frac{\pi}{4}$.

This indicates that the maximum length of a rectangle with a width of $1 \mathrm{~m}$ that can pass through the corridor is
$$
r_{\max }=6 \sqrt{2}-2<6 \times \frac{3}{2}-2=7(\mathrm{~m}) .
$$

Therefore, the equipment cannot be horizontally moved into the corridor.