If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is `3//4` (b) `1//3` (c) `1//4` (d) `2//3`

Options

A. `2/3`
B. `3/4`
C. `1/3`
D. `1/4`

Answer

A

Explanation

Let, the diameter of the sphere be AE = 2r
Let the radius of cone is x and height is y.

` therefore AD = y `
since, `BD^ 2 = AD * DE `
` rArr x ^ 2 = y ( 2r - y ) " " ` …(i)
volume of cone,
` V = ( 1 ) /( 3 ) pi x ^ 2 y = ( 1 ) /( 3 ) pi y ( 2r - y ) y `
` = ( 1 )/( 3 ) pi ( 2 ry ^ 2 - y ^ 3 ) `
On differentiating w.r.t y, we get
` (dV)/( dy ) = ( 1) /( 3 ) pi ( 4ry- 3y ^ 2 ) `
For maxima and minima , put ` (dV) /( dy ) = 0 `
` rArr ( 1 ) /( 3 ) pi ( 4ry - 3y ^ 2 ) = 0 `
` rArr y ( 4r- 3y ) = 0 `
` rArr y ( 4r - 3y ) = 0 `
`rArr y = ( 4 ) /( 3 ) r, 0 `
Again differentiating, we get
` (d^ 2V) /( dy ^ 2 ) = ( 1 ) /( 3 ) pi ( 4r - 6y ) `
At `y = (4 ) /( 3 ) r, ( d^ 2 V) /( dy ^ 2 ) = ( 1 ) /( 3 ) pi ( 4r - 8 r ) = -ve `
` therefore ` Volume of cone is maximum at ` y = ( 4 ) /( 3 ) r `
Now, ratio = ` ("Height of cone")/("Diameter of sphere") `
` = ( y ) /( 2r ) = ( 4r// 3) /(2r ) = ( 2 ) / ( 3 ) `