educationrecreational-mathematics
I have a rectangle of width $20$ meters, length $50$ meters.
How to draw the divider inside this rectangle into $5$ parts, each part has an area of $200m^2$ provided that the total length of the drawn lines inside the rectangle be divided as short as possible.
By request, I'm spinning my comment out into an answer. For $n$ even, say $n=2k$, subdivide the square into a $k\times k$ grid of squares. I'll show it for $k=5$, because I think it's easier to visualize when everything renders as true squares rather than with $\dots$: $$\begin{array}{|c|c|c|c|c|} \hline \,\,&\,\,&\,\,&\,\,&\,\, \\ \hline & & & & \\ \hline & & & & \\ \hline & & & & \\ \hline & & & & \\ \hline \end{array} $$ Now divide it into the top row of squares, the left column of squares, and everything else:
$$\begin{array}{|c|c c c c|} \hline \,\,&\,\,\,\,\mid &\,\,\mid &\,\,\mid& \\ \hline \underline{\,\,\,}& & & & \\ \underline{\,\,\,}& & & & \\ \underline{\,\,\,}& & & & \\ & & & & \\\hline \end{array} $$ (Sorry for the horrible latex hack, but \multicolumn didn't work.) So we have $2k-1$ little squares and one big one, for a total of $2k=n$.
For $n$ odd, split one of the squares into four. (This leaves the "corner case" $n=7$, which I leave as an exercise for the interested reader :)
This is similar to circle packing, where the centers of the circles are the points you want to place in the rectangle. Have a look at this link (second table where $n=4$). Notice how the center circle moves down the more the rectangle deviates from a square.
Using this as inspiration, for your $17 \times 32$ example the second configuration given for $n = 4$ is the way to go. In this configuration the centers of the circles make two equilateral triangles and the max distance $d$ is about $d = 17.19$.
Best Answer
Try to optimize this shape. It should be slightly shorter than 80m
A second option is to use circular arcs that meet at angles of $120^\circ$, and cut the edge at right angles. It gives a total length around 78.25m