A) 4m
B) 3 m
C) 6m
D) 7m
E) 5m
Correct Answer: E
Solution :
\[\because \] \[l=\frac{2\pi r\theta }{360{}^\circ }\] \[\Rightarrow \] \[22=r+r+\frac{2\pi r\theta }{360{}^\circ }\] \[\Rightarrow \] \[\theta =\left( \frac{22-2r}{2\pi r} \right)360{}^\circ \] Now, \[A=\frac{\pi {{r}^{2}}\theta }{360{}^\circ }=r(11-r)\] \[\frac{dA}{dr}=11-2r\]and\[\frac{{{d}^{2}}A}{d{{r}^{2}}}=-2\] \[{{\left( \frac{{{d}^{2}}A}{d{{r}^{2}}} \right)}_{r=\frac{11}{2}}}=-2<0\] \[\therefore \]Area is maximum when\[r=\frac{11}{2}\]. \[\therefore \]Possible radius of circle must be 5 m.You need to login to perform this action.
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