## Task 9 - V00809

For the fencing of a piece of pasture land that is to have a rectangular shape, the LPG New Life has 400 meters of pasture fence available.

At a work meeting, it is suggested to choose the sides of the rectangle to be as equal in length as possible, as the square has the largest area of all rectangles with the same perimeter.

Try to prove this claim! What about the area if the pasture were to be laid out in a circular shape?
If a square is enclosed with $400 \mathrm{~m}$ of fencing

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0565.jpg?height=254&width=800&top_left_y=290&top_left_x=274)
each side of the square will be $100 \mathrm{~m}$ long, and thus the area of the square will be $F_{Q}=100^{2}=10000 \mathrm{~m}^{2}$.

To enclose a rectangle that is not a square, two sides of the square must be extended by $x$ and two sides must be shortened by $x$. (right illustration)

The area of the rectangle is then $F_{R}=(100-x) \cdot(100+x)=10000-x^{2} \mathrm{~m}^{2}$. For any $x>0$, this value is smaller, i.e., $F_{R}<F_{Q}$, which is what we needed to show.

If a circle is enclosed, with a circumference of $400 \mathrm{~m}$, the radius $r=\frac{400 \mathrm{~m}}{2 \pi}$, and thus the area is $F_{K}=\pi\left(\frac{400 \mathrm{~m}}{2 \pi}\right)^{2} \approx 12700 \mathrm{~m}^{2}$. Therefore, the area of a circle is larger than the area of a square with the same perimeter.