Solution:

To begin with, start filling in the details we know and consider drawing some additional lines to look for patterns. In this example, we will look for similar right triangles hidden in the picture. The trick here is to carefully redraw the cylinder/cone as a couple of triangles. Due to the fact they they are similar triangles, we will be able to use the ratio of sides to figure out the \(r\) variable in terms of \(h\): The ratio we can write out is that of similar sides: \[ \solve{ \dfrac{20}{20-h} &=&\dfrac{4}{r} } \] Then we solve this for \(r\) by cross multiplication and careful simplification: \[ \solve{ \dfrac{20}{20-h} &=&\dfrac{4}{r}\\ 20r &=&4(20-h)\\ 5 r &=& 20-h\\ r &=& \dfrac{20-h}{5} } \] We now can use the volume formula for the volume of a cylinder: \[ \solve{ V &=& \pi r^2 h\\ V(h) &=& \pi \left(\dfrac{20-h}{5}\right)^2 h\\ V(h) &=& \dfrac{h\pi\left(20-h\right)^2}{25} } \] We could possible write this out differently, but this is considered simplified enough and answers the question. Here you can visualize how different heights will change the shape of the cylinder: Finally, now that we have the formula for the Volume in terms of \(h\) we can do interesting things like ask ourselves what height of the cone gives the largest volume: It turns out the maximum volume is achieved at \(2/3\) of the total height! Pretty cool stuff.