3. Find the maximum number of rectangles of dimensions $1 \times 10 \sqrt{2}$ that can be obtained from a rectangle of dimensions $50 \times 90$, if cutting along lines parallel to the sides of the given rectangle is allowed.

(Yugoslavia)
3. Let the vertices of the rectangle be $A(0,0), B(90,0), C(90,50)$, and $D(0,50)$. Draw the lines $x+y=10 n \sqrt{2}$ for $n=1,2, \ldots,\left[\frac{90+50}{10 \sqrt{2}}\right]=9$ and consider the segments that the rectangle $A B C D$ intercepts on them. It is easy to calculate that the total length of these segments is $570 \sqrt{2}-360$. However, every rectangle of dimensions $1 \times 10 \sqrt{2}$ with sides parallel to the coordinate axes covers parts of these segments with a total length of $\sqrt{2}$. Therefore, we cannot cut out more than $\left[\frac{570 \sqrt{2}-360}{\sqrt{2}}\right]=315$ of the desired rectangles.

If the given rectangle is divided into rectangles of dimensions $50 \times 60 \sqrt{2}$ and $50 \times (90-60 \sqrt{2})$, we see that the first rectangle can be cut into 300 rectangles of $1 \times 10 \sqrt{2}$, while the second can be cut into an additional $3 \cdot 5=15$ (since $\left[\frac{50}{10 \sqrt{2}}\right]=3$ and $\left[90-60 \sqrt{2}\right]=5$). Thus, it is possible to cut out 315 rectangles.