A rectangular trough is 8ft long, 2ft across the top, and 4 feet deep. if water flows in at a rate of 2 ft 3 / min, how fast is the surface rising when the water is 1 ft deep?

let V be the current volume of the trough and H the current height of the water

then we have

V = 16H

taking derivative in terms of time we get

dV/dt = 16 dH/dt

we know that dV/dt = 2

thus

dH/dt = 2/16 = 1/8

thus the water is rising at

1/8 ft per minute

or

1.5 inches per minute