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I'm going over review problems for a test Wednesday and I'm stuck on this one. Here is the question:

A cylinder with radius 1 and height 3 is sitting inside a cone. Find the dimensions of the cone with minimum volume inside which this cylinder will fit.

This is what I got:

So what I want to find is the radius and height of the cone. This can be represented as a right triangle containing sides ${h=height}$ and ${r=radius}$. There is a similar triangle inside this triangle that has the sides of length ${h-3}$ and ${1}$. Using thales theorem I know that ${h/r = (h-3)/1}$. I solve this for ${r}$ to get ${r=h/(h-3)}$. Now I substitute this value into the volume equation for the cone: ${V=\pi (h/(h-3))^2 h/3 =\pi h^3/3(h-3)^2}$. Now I take the derivative to get ${V'=h^2(h^2-12h+27)/3(h-3)^4}$. Now this where I get stuck. I know that I have to set this to zero and get all the critical points. However, I don't know how to get past ${h(h-12)=-27}$. I know there is probably really simple trick to this but I just cant remember how.

So am I approaching this correctly? Is there a simpler way to solve this? Anny suggestions would be much appreciated. Thanks!

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To solve $h(h-12)=-27$, expand it: $$h^2-12h+27=0\\ \implies h=\dfrac{12\pm\sqrt{144-108}}{2}=\dfrac{12\pm6}{2}=6\pm3=9\text{ or }3$$ Choose the answer that makes physical sense.

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  • $\begingroup$ Ugg I feel so stupid! I don't know why I didn't think of this. $\endgroup$ – boidkan Apr 8 '14 at 3:02

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