The volume of a cylinder is increasing at a rate of 10π cubic meters per hour.

The height of the cylinder is fixed at 5 meters.

At a certain instant, the volume is 80π cubic meters.

What is the rate of change of the surface area of the cylinder at that instant (in square meters per hour) ?

13π/2 m²/h

Step-by-step explanation:

Volume of a cylinder is:

V = πr²h

If h is a constant, then taking derivative of V with respect to time:

dV/dt = 2πrh dr/dt

Surface area of a cylinder is:

A = 2πr² + 2πrh

Taking derivative with respect to time:

dA/dt = (4πr + 2πh) dr/dt

Given that dV/dt = 10π, V = 80π, and h = 5, we need to find dA/dt. But first, we need to find r and dr/dt.

V = πr²h

80π = πr² (5)

r = 4

dV/dt = 2πrh dr/dt

10π = 2π (4) (5) dr/dt

dr/dt = 1/4

dA/dt = (4πr + 2πh) dr/dt

dA/dt = (4π (4) + 2π (5)) (1/4)

dA/dt = 13π/2

The surface area of the cylinder is increasing at 13π/2 m²/h.