An explorer needs to enclose a pentagon with a 255.3 m rope, which is divided by one of its diagonals into a rectangle and an isosceles right triangle, and the area of the pentagon should be as large as possible. Some suggest the diagonal should be 66 m, others suggest 67 m. Which suggestion is better? Is there a better suggestion?
Let the length of the diagonal in question be $x$ (meters), the other side of the rectangle be $z$, so the leg of the isosceles right triangle is $x / \sqrt{2}$, and the perimeter of the pentagon is

$$
2 z + x + 2 \frac{x}{\sqrt{2}} = 255.3 = k
$$

From this,

$$
z = \frac{k - (1 + \sqrt{2}) x}{2}
$$

and the area of the pentagon is

$$
y = x z + \frac{x^2}{4} = \frac{1}{4} \left[2 k x - (1 + 2 \sqrt{2}) x^2 \right]
$$

From this, with $x = 66$, $y = 4255.74 \, \text{m}^2$, and with $x = 67$, $y = 4256.10 \, \text{m}^2$, so the enclosed area is larger when the diagonal length is $67 \, \text{m}$, but the difference is less than $0.4 \, \text{m}^2$. (Taking $\sqrt{2} \approx 1.414213$ because 6 digits are needed).

![](https://cdn.mathpix.com/cropped/2024_05_02_f0455dfaf226def8d5bcg-1.jpg?height=439&width=399&top_left_y=754&top_left_x=855)

A better suggestion can be found by finding the $x$ for which the value of $y$ is maximal. (1) can be written in the form $y = b x - a x^2$, where $b$ and $a$ are positive. By completing the square,

$$
y = -a \left(x^2 - \frac{b}{a} x \right) = -a \left[ \left(x - \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} \right] = \frac{b^2}{4a} - a \left(x - \frac{b}{2a} \right)^2
$$

Here, the second term is negative or 0, so the maximum value of $y$ is $\frac{b^2}{4a}$, and this is achieved when $x = \frac{b}{2a}$. - Now,

$$
b = \frac{k}{2} \quad \text{and} \quad a = \frac{1 + 2 \sqrt{2}}{4} \quad \text{so} \quad x = \frac{k}{2 \sqrt{2} + 1} = \frac{k (2 \sqrt{2} - 1)}{7} \approx 66.69 \, \text{m}
$$

and in this case, the area is

$$
\frac{b^2}{4a} = \frac{k^2 (2 \sqrt{2} - 1)}{28} = \frac{k}{4} \cdot x \approx 4256.19 \, \text{m}^2
$$

which is just about $0.1 \, \text{m}^2$ more than the area obtained with the diagonal $x = 67 \, \text{m}$.

Domokos Zsuzsanna (Makó, József A. g. I. o. t.)

Remarks. 1. The $x$ that gives the largest $y$ can also be found using the inequality between the arithmetic and geometric means. By transformations,

$$
\frac{y}{a} = \frac{b}{a} \cdot x - x^2 = x \left( \frac{b}{a} - x \right)
$$

According to this, the square root of $\frac{y}{a}$ is the geometric mean of the two factors on the right-hand side, but their sum—and thus their arithmetic mean—is constant. Therefore, the product and thus the area of the figure is

$$
x = \frac{b}{a} - x, \quad x = \frac{b}{2a}
$$

maximal when the two factors are equal, and in this case, it is equal to the square of the arithmetic mean.

Mátrai Miklós (Hódmezővásárhely, Bethlen G. g. II. o. t.)

2. For the $x$ that gives the maximum area, $z$ is equal to $x / \sqrt{2}$, i.e., the pentagon has 4 equal sides; in the numerical example, $z = 47.15$ meters.
3. Many contestants tried to solve the problem using differential calculus. We did not accept these because they only concluded that if there is an extremum, it can only be at $x = 66.69$ meters. This is only a necessary condition, but not sufficient.