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If the area of a circular sector of perimeter $60 \mathrm{~m}$ is to be maximized, then its radius must be......... m
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The correct answer is:
$15$
Given, the perimeter of circular sector $=60 \mathrm{~m}$
Let radius be $r \mathrm{~m}$.
Length of $\operatorname{arc} \mathbf{A B}=l$
$\begin{aligned} \therefore \text { perimeter }(p) & =l+2 r \\ l & =p-2 r\end{aligned}$
$\begin{aligned}& \because \text { Area }=\frac{1}{2} l r \\& \Rightarrow A=\frac{(p-2 r) r}{2}=\frac{p r-2 r^2}{2}\end{aligned}$
For $A$ to be maximum, $\frac{d A}{d r}=0$
$\frac{d}{d r}\left(\frac{p r-2 r^2}{2}\right)=0$
$\begin{array}{rlrl}\Rightarrow & & \frac{p-4 r}{2} & =0 \\ \Rightarrow & p & =4 r\end{array}$
$60=4 r$
$\Rightarrow \quad r=15 \mathrm{~m}$
Let radius be $r \mathrm{~m}$.
Length of $\operatorname{arc} \mathbf{A B}=l$
$\begin{aligned} \therefore \text { perimeter }(p) & =l+2 r \\ l & =p-2 r\end{aligned}$
$\begin{aligned}& \because \text { Area }=\frac{1}{2} l r \\& \Rightarrow A=\frac{(p-2 r) r}{2}=\frac{p r-2 r^2}{2}\end{aligned}$
For $A$ to be maximum, $\frac{d A}{d r}=0$
$\frac{d}{d r}\left(\frac{p r-2 r^2}{2}\right)=0$
$\begin{array}{rlrl}\Rightarrow & & \frac{p-4 r}{2} & =0 \\ \Rightarrow & p & =4 r\end{array}$
$60=4 r$
$\Rightarrow \quad r=15 \mathrm{~m}$
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