An ice sculpture in the form of a sphere melts in such a way that it maintains its spherical shape. The volume of the sphere is decreasing at a constant rate of 2pi cubic meters per hour. At what rate, in square meters per hour, is the surface area of the sphere decreasing at the moment when the radius is 5 meters? (Note: For a sphere of radius r, the surface area is 4πr^2 and the volume is 4/3πr^3)

(A) 4pi/5

(B) 40pi

(C) 80pi^2

(D) 100pi

The volume of the sphere is:

V = 4 / 3πr ^ 3

Deriving we have:

V ' = (3) (4/3) (π) (r ^ 2) (r')

From here, we clear the value of r ':

r ' = (V') / ((4) (π) (r ^ 2))

Substituting values:

r ' = (2π) / ((4) (π) ((5) ^ 2))

r ' = (2π) / ((4) (π) (25))

r ' = (2π) / (100π)

r ' = 1/50

Then, the surface area of the sphere is:

A = 4πr ^ 2

Deriving we have:

A ' = 8πrr'

Substituting values:

A ' = 8π (5) (1/50)

Rewriting:

A ' = (40/50) π

A ' = (8/10) π

A ' = (4/5) π

Answer:

The surface area of the sphere is decreasing at:

(A) 4pi / 5