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Topic in related rates

Please show me how do they get the answer 4/9 in/min in this problem:

A solution is passing through a conical filter 24" deep and 16" across the top into a cylindrical vessel of diameter 12". At what rate is the level of the solution in the cylinder rising if when the depth of the solution in the filter is 12" its level is falling at the rate 1"/min?

Please show me how to get the answer in here.. Because I'm really confused how to get this one. Thanks!

Volume of solution in filter paper:
$V = \frac13 \pi r^2 h$

Since radius of cone is 16 and the height 24, the relationship between them is:
16h = 24r
2h = 3r
r = (2h)/3

Substituting this into the equation above;
$V = \frac13 \pi $\frac{2h}{3}$^2 h = \frac{4\pi h^3}{27}$

Now, $\frac{dV}{dh} = \frac{4\pi h^2}{9}$

Using chain rule;
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$

Since the rate of change of the level of the solution is -1 and we have dV/dh

$\frac{dV}{dt} = \frac{4\pi h^2}{9} \times -1 = -\frac{4\pi h^2}{9}$

At height 12, we have $\frac{dV}{dt} = -\frac{4\pi (12)^2}{9} = -64\pi$

Now, this volume goes into the cylindrical vessel with volume.

$V = 2\pi r^2 h$

The volume increase is $64\pi$ in a minute, and r is constant at 12;

$64 \pi = 2\pi (12)^2 h$

Rearranging for h gives:

$h = \frac49$

So, h increases by 4/9 in one minute, its rate is 4/9 in/min.

I hope it helped! :)

thank you very much!^^