The length of a rectangle is three times its width. The perimeter of the rectangle is at most 112 cm.

Which inequality models the relationship between the width and the perimeter of the rectangle?

A. [tex]2w + 2 \cdot (3w) \geq 112[/tex]

B. [tex]2w + 2 \cdot (3w) \textgreater 112[/tex]

C. [tex]2w + 2 \cdot (3w) \textless 112[/tex]

D. [tex]2w + 2 \cdot (3w) \leq 112[/tex]

Let the width be $w$ and the length be $l$. According to the problem, the length is three times the width. Therefore, we have

$$
l = 3w.
$$

The perimeter $P$ of a rectangle is given by

$$
P = 2w + 2l.
$$

Substitute $l = 3w$ into the formula:

$$
P = 2w + 2(3w).
$$

The expression $2w + 2(3w)$ simplifies to

$$
2w + 6w = 8w.
$$

The problem states that the perimeter is at most 112 cm, which means

$$
P \leq 112.
$$

Since $P = 8w$, we can write the inequality as

$$
8w \leq 112.
$$

In the format provided in the answer choices, this inequality corresponds to

$$
2w + 2(3w) \leq 112.
$$

Therefore, the correct inequality is

$$
\boxed{2w + 2(3w) \le 112.}
$$