Given l=3

⇒r2+h2=l2=9
Volume, V=31πr2h
⇒V=31π(9−h2)h=31π(9h−h3)
dhdv=31π(9−3h2)
dhdv=0⇒h=3
dh2d2V=31π(−6h)<0
∴ at h=3 , cone has maximum volume
∴ Vmax=31π(93−33)=23πcm3
JEE Main 2019 — Mathematics Calculus
The maximum volume (incu.m) of the right circular cone having slant height 3m is:
Held on 9 Jan 2019 · Verified 6 Jul 2026.
23π
33π
6π
34π
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