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

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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.

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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`

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Chapter 6: Application of Derivatives - Exercise 6.6 [Page 243]

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NCERT Mathematics Class 12
Chapter 6 Application of Derivatives
Exercise 6.6 | Q 17 | Page 243

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