A) \[\sqrt{p}\]
B) \[\frac{1}{\sqrt{p}}\]
C) \[\frac{p}{2}\]
D) \[\frac{p}{4}\]
Correct Answer: D
Solution :
Perimeter of a sector = p. Let AOB be the sector with radius r. If angle of the sector be q radians, then area of sector \[(A)=\frac{1}{2}{{r}^{2}}\theta \] ?..(i) Length of arc(s) = rq or \[\theta =\frac{s}{r}\]. Therefore perimeter of the sector \[=(p)=r+s+r=2r+s\] ?..(ii) Substituting \[\theta =\frac{s}{r}\] in (i), A = \[\left( \frac{1}{2}{{r}^{2}} \right)\,\left( \frac{s}{r} \right)=\frac{1}{2}rs\] Þ \[s=\frac{2A}{r}\]. Now substituting the value of s in (ii), we get \[p=2r+\left( \frac{2A}{r} \right)\] or \[2A=pr-2{{r}^{2}}.\] Differentiating with respect to \[r,\,\] we get \[2\frac{dA}{dr}=p-4r\]. We know that for the maximum value of area \[\frac{dA}{dr}=0\] or \[p-4r=0\] or \[r=\frac{p}{4}\].You need to login to perform this action.
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