The maximum volume cylinder will be carved when the diameter of sphere and the axis of cylinder coincide.

Let h be the height of our cylinder.

r be the radius of the cylinder.

R = 53 radius of the sphere

OL = x

H = 2x

In Δ AOL,

Volume of cylinder V = πr²h

Therefore,

x = - 5 cannot be taken as the length cannot be negative.

At x = 5, we have to check whether maxima exits or not.

Therefore,

At x = 5, is negative. Hence, maxima exits at x = 5.

Therefore, at x = 5

Volume of Cylinder = π(75 - x²)(2x)

Volume = π(75 - 25)(10) = 500π