18) Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is

\frac{4}{27}\pi h ^3 \tan ^2 \alpha
 

Answers (1)


Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h  respectively

Volume of cylinder = \pi r^2 h'
Volume of cone = \frac{1}{3}\pi R^2 h
Now, we have 
R = h\tan a
Now, since \Delta AOG \and \Delta CEG are similar
\frac{OA}{OG} = \frac{CE}{EG}
\frac{h}{R} = \frac{h'}{R-r}
h'=\frac{h(R-r)}{R}
h'=\frac{h(h\tan a-r)}{h\tan a} = \frac{h\tan a- r}{\tan a}
Now,
V = \pi r^ 2 h' = \pi r^2 .\frac{h\tan a-r}{\tan a} = \pi r^2 h - \frac{\pi r^3}{\tan a}
Now,
\frac{dV}{dr}= 2\pi rh- \frac{3\pi r^2}{\tan a} \\ \frac{dV}{dr}=0\\ 2\pi rh- \frac{3\pi r^2}{\tan a} = 0\\ 2\pi rh = \frac{3\pi r^2}{\tan a}\\ r = \frac{2h\tan a}{3}
Now,
\frac{d^2V}{dr^2}= 2\pi h- \frac{6\pi r}{\tan a}
at r = \frac{2h\tan a}{3}
\frac{d^2V}{dr^2} = 2\pi h- 4\pi h < 0 
Hence,  r = \frac{2h\tan a}{3} is the point of maxima 
h' = \frac{h\tan a-r}{\tan a} = \frac{h\tan a- \frac{2h\tan a}{3}}{\tan a}= \frac{1}{3}h
Hence proved 
Now, Volume (V) at h' = \frac{1}{3}h and  r = \frac{2h\tan a}{3}   is
V = \pi r^2 h' = \pi \left ( \frac{2h\tan a}{3} \right )^2.\frac{h}{3}= \frac{4}{27}.\pi h^3\tan^2 a
hence proved 
 

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions