Problem 10.7. A square board $30 \times 30$ was cut along the grid lines into 225 parts of equal area. Find the maximum possible value of the total length of the cuts.
Answer: 1065.

Solution. The total length of the cuts is equal to the sum of the perimeters of all figures, minus the perimeter of the square, divided by 2 (each cut is adjacent to exactly two figures). Therefore, to get the maximum length of the cuts, the perimeters of the figures should be as large as possible.

The area of each figure is $\frac{900}{225}=4$. There are 5 four-cell figures in total:
![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-40.jpg?height=116&width=1028&top_left_y=417&top_left_x=212)

The perimeter of the square is 8, and the perimeter of all other figures is 10. Therefore, the maximum length of the cuts does not exceed ( $225 \cdot 10 - 120$ )/ $2 = 1065$.

It remains to provide an example. From the above, it follows that this answer will be obtained if we cut the $30 \times 30$ square into any four-cell figures except the square. First, fill the $28 \times 30$ rectangle with $1 \times 4$ strips, leaving a $2 \times 30$ rectangle. The $2 \times 24$ rectangle can also be cut into $1 \times 4$ strips. A $2 \times 6$ rectangle will remain empty, which can easily be cut into two figures in the shape of the letter G and one $1 \times 4$ strip, as shown in Fig. 8.

![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-40.jpg?height=765&width=765&top_left_y=938&top_left_x=348)

Fig. 8: to the solution of problem 10.7