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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πr2h [from Eq. ⇒‌‌V(h)‌‌=‌
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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