# 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!

• It might be easier to think of: When will the area of the entire cone be $9/8$ of the area of the top section of the cone. Then you will be comparing the areas of similar figures, which go as the square of the height. Nov 6 '15 at 16:15
• So $\pi r(r+\sqrt{h^2 + r^2}) = 9/8 Area_{top}$ Where the area of the top is expressed by $\pi r_2 (r_2 +\sqrt{(h-x)^2 +r_2^2})$ Another problem is how do i find the new value of $r$ ($r_2$) in terms of $h_2$ ($h-x$)? Nov 6 '15 at 16:22
• x @Brayden: Note that $h$ and $r$ scale by the same factor (that's how similarity works), and $$\pi ar(ar+\sqrt{(ah)^2+(ar)^2}) = a^2\cdot \pi r(r+\sqrt{h^2+r^2})$$ Nov 6 '15 at 16:28
• Is $a$ some unknown variable related to the scaling of the cone or is it area of a part? Nov 6 '15 at 16:32
• x @Bray: $a$ is whatever you scale one of the cones by to get the other. Nov 6 '15 at 16:33

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.

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

• I have in fact tried that, however i seem to continue to get myself in a mess. I don't exactly know how to express the area bound by the imaginary ring in terms of $r$ and $h$. The cone segment analogy is indeed correct! Nov 6 '15 at 16:13
• See edit for extra tip Nov 6 '15 at 16:15
• See edit for solution Nov 6 '15 at 16:54
• Well damn, good job, thanks heaps @zickens Nov 6 '15 at 17:05
• No problem @Brayden :), when Henning told you about that "scale" factor, he was remarking the fact that the ratios height and base remain the same for this particular problem; whenever you see that the measures of a figure change, but the proportions between the different measures don't, then you can say that the figure has been "scaled". Nov 6 '15 at 17:12