Answer :


Radius of the sphere is .

Volume of cylinder is maximum.

Let us consider,

The radius of the sphere be ‘R’ units.

Volume of the inscribed cylinder be ‘V’.

Height of the inscribed cylinder be ‘h’.

Radius of the cylinder is ‘r’.

Now let AC2 = AB2 + BC2, here AC = 2R, AB =2r, BC = h,

So 4R2 = 4r2 + h2

----- (1)

Let us consider, the volume of the cylinder:

V = πr2h

Now substituting (1) in the volume formula,

---- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with h and then equating it to zero. This is because if the function V(h) has a maximum/minimum at a point c then V’(c) = 0.

Differentiating the equation (2) with respect to h:

[Since ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

3h2π = 4R2π

h = 10

[as h cannot be negative]

Now to check if this critical point will determine the maximum volume of the inscribed cone, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with h:

----- (4)

[Since ]

Now let us find the value of

As , so the function V is maximum at h=10

Substituting h in equation (1)

As V is maximum, substituting h and r in the volume formula:

V = π (50) (10)

V = 500π cm3

Therefore when the volume of a inscribed cylinder is maximum and is equal 500π cm3

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