Question

Twenty meters of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in $sq.m$) of the flower-bed, is

• A. $10$
• B. $25$
• C. $30$
• D. $12.5$

Answer 1

The correct option is B

$25$

Explanation for the correct option:

Step 1: Find the area of the given sector.

In the question, it is given that twenty meters of wire are available for fencing off a flowerbed in the form of a circular sector.

Assume that, the angle of the sector is $\theta$ and the radius of the sector is $r$.

According to the question, $2r+r\theta =20$.

Find $\theta$ in terms of $r$.

$\theta =\frac{20-2r}{r}...1$

Find the area $A$ of the circular sector:

$$\begin{gathered} A=\frac{\theta }{2\mathrm{\pi }}\left({\mathrm{\pi r}}^2\right) \\ \Rightarrow A=\frac{\theta }{2}\left(r^2\right) \end{gathered}$$

Substitute the value of $\theta$ from equation $1$.

$$\begin{gathered} A=\frac{\left(20-2r\right)}{2\mathrm{r}}\left({\mathrm{\pi r}}^2\right) \\ \Rightarrow A=\frac{\left(20-2r\right)r}{2} \\ \Rightarrow A=\left(10-r\right)r \\ \Rightarrow A=10r-r^r...2 \end{gathered}$$

So, the area of the given sector can be given by: $A=10r-r^2$.

Step 2: Find the maximum area of the sector.

Since the area of the given sector is $A=10r-r^2$.

Differentiate both sides with respect to $r$.

$\frac{dA}{dr}=10-2r$

Put $\frac{dA}{dr}=0$ to find the critical point.

$$\begin{gathered} 10-2r=0 \\ \Rightarrow r=5 \end{gathered}$$

Therefore, the value of $r$ for which the area of the given sector is maximum is $5m$.

Substitute $r=5$ in equation $2$.

$$\begin{gathered} A_{max}=10\left(5\right)-{\left(5\right)}^2\Rightarrow A_{max}=50-25 \\ \Rightarrow A_{max}=25 \end{gathered}$$

Therefore, the maximum area of the given sector is $25m^2$.

Hence, option $B$ is the correct option.