13.447 Comparing two bars having the shape of a rectangular parallelepiped, establish that the length, width, and height of the second bar are respectively 1 cm larger than those of the first bar, and the volume and total surface area of the second bar are respectively $18 \mathrm{~cm}^{3}$ and $30 \mathrm{~cm}^{2}$ larger than those of the first. What is the total surface area of the first bar
Solution. Let $x, y, z$ be the linear dimensions of the first block. Then $x+1$, $y+1, z+1$ are the linear dimensions of the second. By the condition

$$
\begin{aligned}
& \left\{\begin{array}{l}
x y z+18=(x+1)(y+1)(z+1) \\
2(x y+y z+x z)+30=2[(x+1)(y+1)+(y+1)(z+1)+(x+1)(z+1)]
\end{array} \Rightarrow\right. \\
& \Rightarrow\left\{\begin{array}{l}
17=x+y+z+x y+x z+y z, \\
6=x+y+z
\end{array} \Rightarrow x y+x z+y z=11, \text { where } 2(x y+x z+y z)=22-\right.
\end{aligned}
$$

the total surface area of the first block.

Answer: $22 \mathrm{~cm}^{2}$.