A) 20
B) 5
C) 15
D) 10
Correct Answer: C
Solution :
Perimeter of sector\[=2r+r\theta \] \[\Rightarrow \] \[60=2r+r\theta \] (given) \[\Rightarrow \] \[\theta =\frac{60-2r}{r}\] Area of sector, \[A=\frac{\pi {{r}^{2}}\theta }{360{}^\circ }\] \[=\frac{\pi {{r}^{2}}(60-2r)}{r\,360}\] \[=\frac{\pi r}{180}(30-r)\] \[\Rightarrow \] \[\frac{dA}{dr}=\frac{\pi }{180}(30-2r)\] For maximum area, \[\frac{dA}{dr}=0\] \[\Rightarrow \] \[30-2r=0\] \[\Rightarrow \] \[r=15\] \[\therefore \] \[\frac{{{d}^{2}}A}{d{{r}^{2}}}=\frac{\pi }{180}(0-2)=\frac{-\pi }{90}<0\] \[\therefore \]It is maximum at\[r=15\text{ }m\].You need to login to perform this action.
You will be redirected in
3 sec