15. (16 points) In an activity class, the teacher asked each student to make a rectangular box with a lid as shown in Figure 6. The length, width, and height of the rectangular box are $x \mathrm{~cm}, y \mathrm{~cm}$, and $z \mathrm{~cm}$, respectively. When Xiao Yang presented his box, he told the class: “The length, width, and height of the box I made are all positive integers, and upon measurement, they satisfy $x y=x z+3, y z=x y+x z-7$." Please calculate how many square centimeters of cardboard are needed to make such a box (ignoring the seams)?
Three, 15. Since $x y=x z+3$, therefore, $x(y-z)=3$.
Also, $x, y, z$ are all positive integers, then
$$
\left\{\begin{array}{l}
x = 3, \\
y - z = 1
\end{array} \text { or } \left\{\begin{array}{l}
x=1, \\
y-z=3 .
\end{array}\right.\right.
$$
(1) When $\left\{\begin{array}{l}x=3, \\ y=z+1\end{array}\right.$, by $y z=x y+x z-7$, that is
$$
(z+1) z=3(z+1)+3 z-7,
$$

rearranging gives
$$
z^{2}-5 z+4=0 .
$$

Thus, the surface area of the rectangular box is
$$
\begin{array}{l}
2(x y+x z+y z) \\
=2(3 \times 2+3 \times 1+2 \times 1)=22\left(\mathrm{~cm}^{2}\right),
\end{array}
$$

or $2(x y+x z+y z)$
$$
=2(3 \times 5+3 \times 4+5 \times 4)=94\left(\mathrm{~cm}^{2}\right) \text { (discard). }
$$
(2) When $\left\{\begin{array}{l}x=1, \\ y=z+3\end{array}\right.$, by $y z=x y+x z-7$, that is
$$
(z+3) z=z+3+z-7,
$$

rearranging gives
$$
z^{2}+z+4=0 .
$$

This equation has no real solutions.
Therefore, to make such a box, at least $22 \mathrm{~cm}^{2}$ of cardboard is needed.