Exercise 4. A farmer owns a square meadow with a side length of $33 \mathrm{~m}$, fenced on all its perimeter. He wishes to divide it into three plots of equal area. Is such a division possible with

1. at most $55 \mathrm{~m}$ of fencing,
2. at most $54 \mathrm{~m}$ of fencing,
Solution: With $33 \text{ m}$ of fencing, one can obtain a plot of $11 \text{ m} \times 33 \text{ m}$ and a plot of $22 \text{ m} \times 33 \text{ m}$. With $22 \text{ m}$ of fencing, the second plot can be divided into two parts of equal area.

Instead of dividing the field into three rectangles, it can be divided into two trapezoids of $16.5 \text{ m}$ height (the sum of the parallel sides being $44 \text{ m}$) and a pentagon, as shown in the figure. Thus, instead of using $22 \text{ m} + 2 \times 16.5 \text{ m} = 55 \text{ m}$, it is sufficient to use $20 \text{ m} + 2 \times \sqrt{16.5^2 + 4^2} \text{ m}$, according to the Pythagorean theorem. Now, $16.5^2 + 16 < 17^2$, because $17^2 - 16.5^2 = (17 - 16.5)(17 + 16.5) = 16.75$, which is sufficient.
![](https://cdn.mathpix.com/cropped/2024_06_04_c88404cd1a030c4b75a5g-4.jpg?height=506&width=1104&top_left_y=780&top_left_x=495)

It is a stroke of luck that the trapezoid with parallel sides of $20 \text{ m}$ and $24 \text{ m}$ works. We could have more generally studied the trapezoid with parallel sides $22 - x$ and $22 + x$, which amounts to studying the function defined by $f(x) = 22 - x + 2 \sqrt{16.5^2 + (2x)^2}$. It is for $5/3 < x < 13/5$ that $f(x) < 54$, and in particular for $x = 2$, but the minimum $22 + 33/4 \sqrt{15} = 53.9521126 \ldots$ is reached at $x = 11/5 \sqrt{15} = 2.13014 \ldots$