Among circular segments with the same perimeter, which one has the largest area?
The perimeter of a circular segment is $k=i+2 r$, where $i$ is the arc length, and $r$ is the radius. The area of the circular segment is $t=\frac{r i}{2}$. Substituting the value of $i$ from the perimeter formula into the area formula:

$$
t=\frac{r i}{2}=\frac{r(k-2 r)}{2}=-r^{2}+\frac{k}{2} r
$$

(where $k$ is constant). Thus, the area is a quadratic function of the radius, whose graph is a parabola, and since the coefficient of the quadratic term is negative, the parabola opens "downwards," meaning the function has a maximum.

![](https://cdn.mathpix.com/cropped/2024_05_02_1c2360ec6530e160bfccg-1.jpg?height=412&width=420&top_left_y=477&top_left_x=842)

By completing the square for the quadratic polynomial, we can read off the location and the value of the function's maximum: $t=-\left(r-\frac{k}{2}\right)^{2}+\frac{k^{2}}{4}$. The maximum occurs where the radius is one-fourth of the circular segment's perimeter, at which point $i=2 r$, and the central angle of the circular segment $\alpha=2$ radians $\approx 114.59^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1c2360ec6530e160bfccg-1.jpg?height=211&width=290&top_left_y=1087&top_left_x=907)