In a circle of radius $R$, we draw the chord that, as the circle rotates around a diameter parallel to the chord, generates a cylindrical surface of maximum area.
Let the chord $AB = 2x$, then the surface area of the cylinder's lateral surface is:

$$
y = 2 \sqrt{R^2 - x^2} \cdot \pi \cdot 2x = 4 \pi \sqrt{(R^2 - x^2) x^2}
$$

Since $(R^2 - x^2) + x^2 = R^2 = \text{const.}$, the maximum occurs when:

$$
R^2 - x^2 = x^2 \text{, i.e., } x = \frac{R}{2} \sqrt{2} \text{ and thus } AB = R \sqrt{2}
$$

Therefore, the sought chord is one side of the chord square.

(Hildegárd Szmodics, Kaposvár.)

Number of solutions: 35.