What is height of cylinder where volume is maximum when subscribed inside a hollow sphere of radius r is?

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Solution:

A sphere of fixed radius (R) is given.

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

From the given figure, we have

h = 2 √ (R2 - r2)

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

V = π r2h

= 2π r2 √ (R2 - r2)

Therefore,

V = π r2h

= 2 π r2 √ (R2 - r2)

dV/dr = 4π r√ (R2 - r2) + (2π r2 (-2r)) / 2 √ (R2 - r2)

= 4πr√ (R2 - r2) - (2π r3) / √ (R2 - r2)

= [4π r √ (R2 - r2) - 2π r3] / √ (R2 - r2)

= (4π rR2 - 6π r3) / √ (R2 - r2)

Now,

dV/dr = 0

⇒ (4π rR2 - 6π r3) / √ (R2 - r2) = 0

4π rR2 = 6π r3

r2 = 2R2/3

Also,

d2V / dr2 = {√ (R2 - r2) (4π R2 - 18π r2) - (4π rR2 - 6π r3) ((- 2R) / 2 √ (R2 - r2))} / (R2 - r2)3/2

Now, it can be observed that at r2 = 2R2/3,

⇒ d2V / dr2 < 0

Thus, the volume is the maximum when r2 = 2R2/3

when, r2 = 2R2 / 3

Then, the height of the cylinder is

h = 2√ R2 - 2R2 / 3

= 2 √ (R2 / 3)

= 2 R / √ 3

Hence, the volume of the cylinder is the maximum when the height of the cylinder is 2 R / √ 3.

Now finding the maximum volume of the cylinder:

V = π hR2 - (π h3/4)

Replacing h with 2 R / √ 3, we get

V = (2π R3 / √ 3) - (π / 4 x 8R3/3√ 3)

V = (2π R3 / √ 3) - (2 π R3/3√ 3)

V = (2π R3 / √ 3) × 2/3

V = (4π R3 / 3√ 3)

Hence the maximum volume will be (4π R3 / 3√ 3) cubic units

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 17

Summary:

Hence we have shown that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. The maximum volume will be (4π R3 / 3√ 3) cubic units

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The first question that comes into my mind here is whether any cylinder that touches(at 4 pts) the circumference of the sphere and does not go out of it, has equal volume?

Second, how do i mathematically limit the volume of the cylinder to be less than that of a sphere? Squeeze theorem?

Please help, thanks!