A) \[3\sqrt{3}\pi \]
B) \[6\pi \]
C) \[2\sqrt{3}\pi \]
D) \[\frac{4}{3}\pi \]
Correct Answer: C
Solution :
| Volume of cone \[\text{V}\,=\,\frac{1}{3}\pi {{r}^{2}}\,h\] |
| \[and\,\,{{r}^{2}}+{{h}^{2}}\,=\,{{l}^{2}}\,where\,\,l\,=\,3\] |
| \[\Rightarrow \] \[\text{V}\,=\,\frac{\pi }{3}\,(9-{{\text{h}}^{2}})\,\text{h}\] |
| \[\frac{dV}{dh}\,=\,0\] |
| \[\Rightarrow \] \[\text{h}\,=\,\sqrt{3}\] |
| and \[\left( \frac{{{d}^{2}}V}{d{{h}^{2}}} \right)\,=\,-\,6h\,<\,0\] |
| \[\Rightarrow \] V is max, at h \[=\,\sqrt{3}\] max. \[V\,=\,\frac{\pi }{3}\,(9-3)\sqrt{3}\,=\,2\sqrt{3}\pi \,\text{m}.\] |
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