question_answer

The maximum volume (in cu. m) of the right circular cone having slant height 3 m is: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

A) \[6\pi \]

B) \[\frac{4}{3}\pi \]

C) \[2\sqrt{3}\,\pi \]

D) \[3\sqrt{3}\,\pi \]

Correct Answer: C

Solution :

\[V\,\,=\,\,\frac{1}{3}\,\pi {{r}^{2}}h\] (Given \[{{h}^{2}}+\text{ }{{r}^{2}}=\text{ }9\]) \[V\,\,=\,\,\frac{1}{3}\pi \,(9-{{h}^{2}})h\] \[\frac{dV}{dh}\,=\,\frac{1}{3}\,\pi \,\,[9-3{{h}^{2}}]\,\] For maximum & minimum \[\frac{dV}{dh}\,\,\,=\,\,0\] \[9\,\,=\,\,3{{h}^{2}}\] \[{{h}^{2}}=3\] \[\frac{{{d}^{2}}V}{d{{h}^{2}}}\,=\,\frac{1}{3}\,\pi \,\,(-6h)\,\,<\,\,0\,\] Maximum at \[h\,=\,\sqrt{3}\] \[{{r}^{2}}=9-{{h}^{2}}\] \[{{r}^{2}}=9-3\] \[{{r}^{2}}=\text{ }6\] \[{{V}_{max}}\,=\,\,\frac{1}{3}\,\pi \,\,\times \,6\,\,\times \,\,\sqrt{3}\,\,\,=\,\,2\sqrt{3}\pi \,\,\]