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I need to express the radius $r$ of the right circular cone as a function of its height $h$ given that its volume equals to its lateral surface area. I know the two equations $\pi r \times \sqrt{r^2 + h^2}$ and the volume $\frac{\pi}{3} r^2 h$. Do I just set these equations as equal and solve for $h$? I'm not quite sure where to go from here.

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You did not state which geometrical object you deal with... If you need to find $r(h)$, you should be solving for $r$, rather than $h$. – Sasha Nov 1 '11 at 18:33
Sorry, this is a right circular cone. – erimar77 Nov 1 '11 at 18:33
$\sqrt{r^2+h^2}=\frac{1}{3}rh$ , square both sides and solve for $r$ – pedja Nov 1 '11 at 18:37
@erimar77 I have edited your question to state the geometric object under consideration as well as to include the constraint. If you agree with the changes, please remove *[added]* and *[/added]* from the text. – Sasha Nov 1 '11 at 18:54
Am I the only one who really dislikes this question (as homework)? Maybe it's the physicist in me, but volume and area are different notions, and equating the two is inherently unit-dependent in a painfully ungeometric way. I appreciate the need to introduce algebraic questions in some fashion, but surely that can be done in a way at least somewhat less arbitrary than this... – Steven Stadnicki Nov 1 '11 at 19:01
up vote 1 down vote accepted

The lateral surface area is $A_L = \frac{1}{2} C \times S$, where $C$ is the base circumference, and $S = \sqrt{r^2+h^2}$ is the distance from the tip of the cone to the point on the base circle.

Equality of the volume and the surface area gives you an equation: $$ \pi r \sqrt{r^2+h^2} = \frac{\pi}{3} r^2 h $$ Assuming $r>0$ and $h>0$, this simplifies to $3 \sqrt{r^2+h^2} = r h$. Squaring the left-hand-side and the right-hand-side will give the auxiliary equation, with the property that every solution of the original equation being a solution of the auxiliary equation. But the auxiliary equation might have extraneous solutions.

The auxiliary equation will be a simple quadratic equation, with two solutions. You should check which one of these will satisfy the original equation, and under which conditions this will be possible.

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