Optimization of a split cone This was a thought that I've been pondering on for several days now, however I can not seem to optimize using calculus correctly in relation to the following statement. Any help would be greatly appreciated.
Assume you have 3 dimensional cone with a perfectly circular base.
Let the height of the cone be $h$ and the radius $r$.
Now, a certain distance $x$ from the base, there is an imaginary ring drawn parallel to that of the base.
Assume where all other statements involving "rest of the area" to be the whole surface area of the cone above that of the imaginary ring $x$ from the base.
So, when will the area of the surface bound between the base and the imaginary ring be equal to one eighth the rest of the surface area?
Thanks, this should put my mind at ease!
 A: The problem itself is not very difficult. I still don't have enough reputation to comment, so I'm going to suggest this as an answer: define surface area as a function of $h$ and use pythagoras' theorem; also, consider that if, and I'm taking a guess here because it is unclear in your statement, the ring touches the edge of the cone, your original figure gets divided into a "cone segment" below, and a new, smaller cone above.

Actual answer:
To solve this problem, you don't actually have to use optimization. All you need to do is remember your algebra.
First, consider that if the cone has height $H$, and radius $R$, then the cone will have a surface area $S_{t} = \pi R^{2} + \pi R \sqrt{H^{2} + R^{2}}$
Now, if we choose a number $x$ between $0$ and $R$ and truncate the cone, we can deduce that 
$$
\frac{H}{R}=\frac{h_{1}}{x}
$$
Where $h_{1}$ is the height of the upper cone created by the truncation of the original cone, simply because $\tan{\alpha} = \frac{H}{R}$ but also $\tan{\alpha} = \frac{h_{1}}{x}$, hence the above.
Now, let us define the surface area $S_{1} = \pi x^{2} + \pi x \sqrt{x^{2} + \left(\frac{xH}{R}\right)^{2}}$, which represents the surface area of the upper cone, and the $S_{2}$ represents the surface area of the lower cone segment truncated by $x$.
But since $S_{2} = \frac{1}{8}S_{t}$, we have
$$
S_{t} = S_{1} + \frac{1}{8}S_{t}
$$
$$
\frac{9}{8}S_{t}=S_{1}
$$
$$
\frac{9}{8}\left(\pi R^{2} + \pi R \sqrt{H^{2} + R^{2}}\right) = \pi x^{2} + \pi x \sqrt{x^{2} + \left(\frac{xH}{R}\right)^{2}}
$$
$$
x=\frac{3R}{2\sqrt{2}}
$$
Computed with WolframAlpha for $x$, $R$ and $H$ positive
