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Question
Show that the height of the cylinder of maximum volume, that can be inscribed in a sphere of radius R is `(2R)/sqrt3.` Also, find the maximum volume.
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Solution 1
Given that the radius of the sphere is R,.
Let r and h be the radius and 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 at its maximum when the height of the cylinder is `(2R)/3`
Solution 2
Let the radius of the sphere, OA = R
makes an angle θ with the axis of the cylinder.
Radius of cylinder = R sin θ
Height of cylinder = 2R cos θ
∴ Volume of cylinder = πr2h
V = π (R sin θ)2 × 2 R cos θ
= 2πR3 sin2 θ cos θ
On differentiating with respect to θ,
`(dV)/(d theta) = 2piR^2 [sin^2 theta (- sin theta) + cos theta * 2 sin theta cos theta]`
= 2πR3 [- sin3 θ + 2 cos2 θ sin θ]
= 2πR3 sin θ (2 cos2 θ - sin2 θ)
= 2πR3 sin θ (2 cos2 θ - 1 + cos2 θ)
= 2πR3 sin θ (3 cos2 θ - 1)
For maximum and minimum, `(dV)/(d theta) = 0`
⇒ 2πR3 sin θ (3 cos2 θ - 1) = 0
3 cos2 θ - 1 = 0 या `cos^2 theta = 1/3`
`therefore cos theta = 1/sqrt3`
At `cos theta = 1/sqrt3` the sign of `(dV)/(d theta)` changes from positive to negative when θ passes through cos θ = `1/sqrt3`.
V is maximum at `=> cos theta = 1/sqrt3`.
Height = 2 R cos θ = 2R `* 1/sqrt3 = (2R)/3`
∴ Maximum volume of cylinder = 2πR3 sin2 θ cos θ
`= 2piR^3 (sqrt2/sqrt3)^2 1/sqrt3 ...[because cos theta = 1/sqrt3, sin theta = sqrt2/sqrt3]`
`= 2piR^3 xx 2/3 * 1/sqrt3`
`= (4 piR^3)/(3 sqrt3)` square unit.
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