Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the…

1. $12.5$
2. $10$
3. $25$
4. $30$

Solution

Perimeter $=2r+r\theta =20$



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

Now, Area $A=\frac{r^2\theta }{2}$

$\Rightarrow A=\frac{r^2}{2} \left(\frac{20-2r}{r}\right)=10r-r^2$

$\Rightarrow \frac{dA}{dr}=10-2r=0$

$\Rightarrow r=5$

${\frac{d^{2}A}{d{r}^2}}_{\left(r=5\right)}=-2<0\Rightarrow r=5$ is the point of maxima

$A_{max}=10\times 5-25=25$

Asked in: JEE Main 2017 (02 Apr)

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