• This is a related rates problem. The key is to find a relation (read, "equation") between the volume V and the height of the pile, h. Since we're trying to find how fast the height is increasing, we're looking for h', the first time-derivative of h. Since the pile is a right, circular cone whose base diameter is equal to the height, we can write the volume of the cone as follows: V = (1/3)*(pi)*r^2*h = (pi/3)*(h/2)^2*h = (pi/12)*h^3 since the the diameter D = 2r = h. Since we're looking for h', and we're given h=24ft and V'=20ft^3/min, a good idea would be to implicitly differentiate this equation since it'll yield a differential equation with everything we need to solve for h'. So we implicitly differentiate to get V' = (pi/12)*(3*h^2*h') = (pi/4)*h^2*h' Which means that we can solve for h': h' = 4*V'/(pi*h^2) Now we just plug in V'=20 and h=24 to get h' = 4*(20)/(pi*(24)^2) = 5/(36*pi) which is about equal to .0442097 ft/min. This makes sense because it's positive, which means the pile is getting bigger, which we would expect since V' is positive.

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