## Task A-2.6.

A wire of length $d$ is to be used to enclose a plot of land in the shape of a circular sector so that the area of this plot is maximized. What is the area of the enclosed plot of land?
## Solution.

Let $r$ be the radius of the circular sector, and $l$ the length of its arc. Then $2 r + l = d$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5bba13a74780d91c8bdbg-15.jpg?height=682&width=739&top_left_y=550&top_left_x=590)

The area of the circular sector is $P = \frac{r l}{2}$.

Therefore, $P = \frac{r l}{2} = \frac{1}{2} r(d - 2 r)$.

Notice that the dependence of $P$ on $r$ is given by the quadratic function $P(r) = -r^2 + \frac{1}{2} d r$. The maximum value of this function is achieved at $r = \frac{d}{4}$.

By substituting $r = \frac{d}{4}$, we find that the maximum possible area is $\frac{d^2}{16}$.

Note: The formula for the area of a circular sector $P = \frac{r l}{2}$ does not need to be derived by the student. However, if the student does not know this formula, it can be derived as follows.

Let $\alpha$ be the measure of the angle subtended by the two radii of the circular sector. Then the area of the circular sector is $P = \frac{\alpha}{360^\circ} \cdot r^2 \pi$, and the length of the arc of the circular sector is $l = \frac{\alpha}{360^\circ} \cdot 2 r \pi$. By eliminating the variable $\alpha$, we obtain the desired formula.

Note: The maximum possible value of the area can also be determined using the inequality between the arithmetic and geometric means. If we apply this inequality to the positive numbers $r$ and $\frac{d}{2} - r$, we see that

$$
P = r \left( \frac{d}{2} - r \right) \leqslant \left( \frac{r + \frac{d}{2} - r}{2} \right)^2 = \frac{d^2}{16}
$$

In this approach, it is not necessary to comment that the maximum possible value is achieved at $r = \frac{d}{4}$ (although this is easily seen because equality holds if and only if $r = \frac{d}{2} - r$).