Chow fast is the height changing when the volume of beer in the fermenter is 486?

Question

A beer fermenter is in the shape of an inverted cone it is being filled at a rate of 40ft cube per min. The fermenter has a height of 100 ft and a radius at the top of 50 feet. This means the radius is always half of the height. How fast is the height changing when the volume of beer in the fermenter is 486pie ft square per min?

Answer 1

Volume of cone:

 

     V = pi·r²·h/3

 

Since r = h/2:

 

     V = pi·(h/2)²·h/3 = pi·h³/12

     dV/dt = pi·3h³/12 · dh/dt = pi·h³/4 · dh/dt

 

Solving for dh/dt:

 

     4·(dV/dt)/(pi·h²) = 200/(pi·h²) = dh/dt        (since dV/dt is given as 50 ft3/min)

 

Now we just need the value of h².  Going back to the volume equation:

 

     V = pi·h³/12

     486·pi = pi·h³/12    (V is given as 486pi)

     12·486 = h³

     5832 = h³

     (5832)^2/3 = (h³)^2/3

     324 = h²

 

200/(pi·h²) = dh/dt

200/(324·pi) = dh/dt

50/(81pi) ≈ 0.2 ft/min = dh/dt