Advertisement Remove all ads

Show that the Height of the Cylinder of Maximum Volume, Which Can Be Inscribed in a Sphere of Radiu - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

Show that the height of the cylinder of maximum volume, which can be inscribed in a sphere of radius R is `(2R)/sqrt3.`  Also find the maximum volume.

Advertisement Remove all ads

Solution

Given, radius of the sphere is R.

Let r and h be the radius and the height of the inscribed cylinder respectively.

From the given figure, we have `h=2sqrt(R^2-r^2)`

The volume (V) of the cylinder is given by,

`V=pir^2h=2pir^2sqrt(R^2-r^2)`

`therefore (dV)/(dr)=4pirsqrt(R^2-r^2)+(2pir^2(-2r))/(2sqrt(R^2-r^2))`

`=(4pirR^2-6pir^3)/sqrt(R^2-r^2)`

for maxima or minima, `(dV)/(dr) =0 `

`(4pirR^2-6pir^3)=0`

`r^2=(2R^2)/3`

Now `,(d^2V)/(dr^2)=(sqrt(R^2-r^2)(4piR^2-18pir^2)-(4piR^2-6pir^3)(-2r)/(2sqrt(R^2-r^2)))/(R^2-r^2)`

`=((R^2-r^2)(4piR^2-18pir^2)-r(4piR^2-6pir^3))/(R^2-r^2)^(3/2)`

`=(4piR^4-22pir^2R^2+12pir^4+4pir^2R^2)/(R^2-r^2)^(3/2)`

Now, it can be observed that when `r^2=(2RR62)/3,(d^2V)/(dr^2)<0`

∴The volume is the maximum when `r^2=(2R^2)/3`

Maximum volume = V = `pih((4R^2 - h^2)/4)`
            `h = 2R/sqrt3`

`V_(max) = pi xx 2R/sqrt3 ((4R^2 - 4R^2/3)/4)`

` = (2piR)/sqrt3 . (2R^2)/3 = (4piR^3)/(3sqrt3)` cubic units

Hence, the volume of the cylinder is the maximum when the height of the cylinder is `(2R)/3`

Concept: Maxima and Minima
  Is there an error in this question or solution?

APPEARS IN

NCERT Class 12 Maths
Chapter 6 Application of Derivatives
Q 17 | Page 243

Video TutorialsVIEW ALL [2]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×