## Task B-3.4.

What is the maximum possible area of a circular sector whose perimeter is $40 \mathrm{~cm}$?
## First Solution.

The perimeter of a circular sector with radius $r$, central angle $\alpha$, and corresponding arc length $l$ is $o=2 r+l=2 r+\frac{r \alpha \pi}{180^{\circ}}=40$.

First, let's express the measure of the angle $\alpha$ from the obtained equality.

$\alpha=\frac{180^{\circ}}{r \pi}(40-2 r)$

Now we can express the area of the circular sector as a function of its radius $r$.

$P=\frac{r^{2} \alpha \pi}{360^{\circ}}=\frac{r^{2} \pi}{360^{\circ}} \cdot \frac{180^{\circ}}{r \pi}(40-2 r)=\frac{r}{2}(40-2 r)=20 r-r^{2}$

Notice that $P(r)$ is a quadratic function of $r$ which reaches its maximum value at $r_{0}=-\frac{20}{2 \cdot(-1)}=10 \mathrm{~cm}$.

The maximum value of the function $P$, or the largest possible area of the circular sector with the given perimeter, is: $P(10)=20 \cdot 10-10^{2}=200-100=100 \mathrm{~cm}^{2}$.