4. To cut two different sizes of steel plates into three specifications $A$, $B$, and $C$, each type of steel plate can simultaneously cut the number of smaller steel plates of the three specifications as shown in the table below:
\begin{tabular}{|l|c|c|c|}
\hline & $A$ specification & $B$ specification & $C$ specification \\
\hline The first type of steel plate & 2 & 1 & 1 \\
\hline The second type of steel plate & 1 & 2 & 3 \\
\hline
\end{tabular}

Now, 15, 18, and 27 pieces of specifications $A$, $B$, and $C$ are required, respectively. The number of sheets of the two types of steel plates required are $m$ and $n$ ($m, n \in \mathbf{N}$), then the minimum value of $m+n$ is \qquad .
$3 y=27$ intersects at $\left(\frac{18}{5}, \frac{39}{5}\right)$, so $m+n \geqslant \frac{18}{5}+\frac{39}{5}=\frac{57}{5}$, and when $m=4, n=8$ we have $m+n=12$. Therefore, the minimum value of $m+n$ is 12.