A wire of length \(20 \mathrm{~cm}\) is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is

Given that
length of the wire
\[
\mathrm{P}=20 \mathrm{~cm}
\]
Then, \(\mathrm{P}=\) diameter + arc length
\[
\begin{aligned}
20 &=2 r+S \\
S &=20-2 r \\
S &=2(10-r) \quad \text{...(i)}
\end{aligned}
\]
Also, know that area of semicircle
\[
\begin{aligned}
& \mathrm{A} =\frac{1}{2} \pi \mathrm{r}^{2} \quad \text{...(ii)} \\
\Rightarrow & \mathrm{A} =\frac{1}{2}(\pi \mathrm{r})(\mathrm{r}) \\
\because & \text { Angle } =\frac{\mathrm{Arc}}{\text { Radius }} \\
\Rightarrow & \pi =\frac{\mathrm{S}}{\mathrm{r}}
\end{aligned}
\]
\(\Rightarrow S=r \pi\) for straight length of wire
\[
\Rightarrow \quad \mathrm{A}=\frac{1}{2} \mathrm{~S} \cdot \mathrm{r} \quad \text{...(iii)}
\]
From Eq. (i)
\[
A=\frac{1}{2} \cdot 2(10-r) \cdot r
\]
\[
A=10 r-r^{2} \quad \text{...(iv)}
\]
Now, \(\quad \frac{\mathrm{dA}}{\mathrm{dr}}=10-2 \mathrm{r}\)
For max or min area of enclosed by wire
\[
\frac{\mathrm{dA}}{\mathrm{dr}}=0 \Rightarrow 10-2 \mathrm{r}=0
\]
\(\Rightarrow \quad \mathrm{r}=5\)
Then, from Eq. (iv)
\[
\begin{aligned}
&A=10(5)-(5)^{2} \\
&A=50-25 \\
&A=25 \mathrm{sq} \mathrm{cm}
\end{aligned}
\]