JEE Mains · Maths · STD 12 - 6. Application of derivatives
The maximum volume (in \(cu.m\)) of the right circular cone having slant height \(3\, m\) is
- A \(6\pi \)
- B \(3\sqrt {3\,} \pi \)
- C \(\frac{4}{3}\,\pi \)
- D \(2\sqrt {3\,} \pi \)
Answer & Solution
Correct Answer
(D) \(2\sqrt {3\,} \pi \)
Step-by-step Solution
Detailed explanation
\(t=3\) \(r^{2}+h^{2}=9\) Volume of cone is \(=\frac{1}{3} \pi r^{2} h\) \(\mathrm{V}=\frac{1}{3} \pi \mathrm{h}\left(9-\mathrm{h}^{2}\right)\) \(\frac{d v}{d h}=\frac{1}{3} \pi\left(9-3 h^{2}\right)=0\) \(9-3 h^{2}=0\) \(h^{2}=3, h=\sqrt{3}\) \(V=\frac{1}{2}(\pi)(6) \sqrt{3}\)…
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