A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[{{4}^{C}}\]
D) \[{{2}^{C}}\]
Correct Answer: D
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\[\theta ={{2}^{o}}\]
You need to login to perform this action.
You will be redirected in
3 sec
