Application of Derivatives Part 1

Let h= height of the cone,
r= radius of circular base =√(3)2−h2‌‌‌‌‌‌‌[∵l2=h2+r2]
=√9−h2....(i) Now, volume (V) of cone =‌

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3

πr2h
[from Eq. ⇒‌‌V(h)‌‌=‌

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3

π(9−h2)h
‌‌=‌

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π[9h−h3]
For maximum volume V′(h)=0 and V′′(h)<0.
Here, ‌‌V′(h)=0⇒(9−3h2)=0
⇒‌‌h=√3   [∵h≮0] and ‌‌V′′(h)=‌

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π(−6h)< Ofor h=√3
Thus, volume is maximum when h=√3 Now, maximum volume
V(√3)‌‌=‌

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π(9√3−3√3) ........ from Eq. (ii)
‌‌=2√3π