logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Show that volume of the greatest cylinder which can inscribed in a cone of height h and semi-vertical angle $ 30^{\circ} \: is \: \large\frac{4}{81} $$\: \pi h^3.$

Can you answer this question?
 
 

2 Answers

0 votes

Let R and H be the radius and height of the cylinder.

Given that r and h are the radius and height of the cone respectively

image

image

image

image

image

image

image

Differentiate with respect to R

image

For Maxima or minima image

image

image

image

image

Diff image with respect to R

image

image

image

image

V is maximum when image

image

image

image

image

image

answered Dec 26, 2014 by rameshmathematic
 
0 votes
 
Best answer

Let the height of the cylinder be x, and height of the cone be h

 Therefore $VO' = xcot30° = x\sqrt{3}$

therefore $ OO' = h - x\sqrt{3} $

V= \pi x^2(h-x\sqrt{3}) $

on differentiating w.r.t x we get,

dv/dX = 2πxh - 3πx^2√3

For maximum or minimum V 

dv/dX = 0

Therefore 2πxh - 3πx^2√3 = 0

This implies x = (2h/3√3)

Again differentiating w.r.t x we get,

d^2y/dx^2 = 2πh - 6πx√3

Substituting for x we get,

d^2y/dx^2 = π(2h- 4h)

Hence V is maximum when x = 2h/3√3

The maximum volume of the cylinder is given by 

V = π[(2h/3√3)^2(h- 2h/3)]

V = (4/81)×πh^3

Hence proved.

answered Dec 29, 2014 by vijayalakshmi.r
selected Dec 29, 2014 by pady_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...