question_answer

The perimeter of a sector is p. The area of the sector is maximum when its radius is                                         [Karnataka CET 2002]

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 \[r=\frac{p}{4}\].