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Q.
The perimeter of a sector is constant. If its area is to be maximum, then sectorial angle is
Jharkhand CECEJharkhand CECE 2015
Solution:
Let $ r $ be the radius of the circle and $ \theta $ be the sectorial angle of a sector of it. Then, Perimeter $ =2r+r\theta =k $ (constant) [given]
$ \Rightarrow $ $ r=\frac{k}{2+\theta } $
Let $ A $ be the area of the sector, then
$ A=\frac{1}{2}{{r}^{2}}\theta =\frac{{{k}^{2}}}{2}\cdot \frac{\theta }{{{(\theta +2)}^{2}}} $
On differentiating both sides, w.r.t. $ \theta $ , we get
$ \frac{dA}{d\theta }=\frac{{{k}^{2}}}{2}\left\{ \frac{{{(\theta +2)}^{2}}-2\theta (\theta +2)}{{{(\theta +2)}^{4}}} \right\} $
$ =\frac{{{k}^{2}}}{2}\frac{(2-\theta )}{{{(\theta +2)}^{3}}} $
For maximum, put $ \frac{dA}{d\theta }=0 $
$ \Rightarrow $ $ \theta =2 $
Now, $ \frac{{{d}^{2}}A}{d{{\theta }^{2}}}=\frac{{{k}^{2}}}{2}\left[ \frac{2\times (-3)}{{{(\theta +2)}^{4}}}-\frac{{{(\theta +2)}^{3}}\times 1-\theta \times 3{{(\theta +2)}^{2}}}{{{[{{(\theta +2)}^{3}}]}^{2}}} \right] $
$ =\frac{{{k}^{2}}}{2}\left[ \frac{-6}{{{(\theta +2)}^{4}}}-\frac{\theta +2-3\theta }{{{(\theta +2)}^{4}}} \right] $
$ =\frac{-{{k}^{2}}}{2}\left[ \frac{6}{{{(\theta +2)}^{4}}}+\frac{2-\theta }{{{(\theta +2)}^{4}}} \right] $ At $ \theta =2 $ ,
$ \frac{{{d}^{2}}A}{d{{\theta }^{2}}}=\frac{{{k}^{2}}}{2}\left[ \frac{6}{{{4}^{4}}}+0 \right] $ $ =\frac{-3{{k}^{2}}}{256}<0 $
Hence, $ A $ is maximum, when $ \theta ={{2}^{o}} $