Best way to cut a cake to avoid dryness

[adimare]

TL;DR

You have a cylindrical cake of radius R, height h. Assuming the outer surface of the cake seals the inside very well, how do you cut a slice of volume 1/N that minimizes the surface area that's exposed to the outside? Once the cake has been cut, you can't reshape or reposition the remaining cake in order to avoid dryness.

There's a very famous letter to the editor of Nature from 1906 that describes a process to cut a cake in a way that minimizes the exposed area to become dry. The solution is to cut a strip of cake from the middle with tho parallel lines, then put together the remaining chords together in order to stop them from getting dry. Here's a link to the letter: https://galton.org/essays/1900-1911/galton-1906-cake.pdf

However, let's say we add a restriction that forbids us from avoiding dryness by putting the remaining pieces of cake together and we actually have to come up with a way to cut a piece of cake that minimizes the surface area exposed to dryness, it becomes non trivial to solve.

Basically:
You have a cylindrical cake of radius R, height h. Assuming the outer surface of the cake seals the inside very well, how do you cut a slice of volume 1/N that minimizes the surface area that's exposed to the outside? Once the cake has been cut, you can't reshape or reposition the remaining cake in order to avoid dryness.

 

[pasmith]

Slicing off the top 1/N exposes a surface area of \pi R^2.
Slicing a sector of angle 2\pi/N radians exposes a surface area of 2Rh.

Which is larger depends on whether \frac{h}{R} > \frac{\pi}{2}.

Alternatively, you can cut a sector of height h&#039; &lt; h and angle \theta such that \frac12 R^2\theta h&#039; = R^2 h \pi/ N, giving a surface area of <br /> 2Rh&#039; + \frac{2\pi h}{Nh&#039;}

 

[Hill]

Slicing off the top $1/N$ reduces the exposed area by $2 \pi Rh/N$.
Slicing a sector of angle $2 \pi /N$ radians, adds $2Rh$ to the exposed area and removes $2 \pi Rh/N + 2 \pi R^2 /N$ from it.
$h \gt \pi R/N$?

P.S. Sorry, I've missed this part:

Assuming the outer surface of the cake seals the inside very well,

 

[adimare]

Slicing off a sector doesn't seem to be a good approach. No matter what the value of $N$ is, you'll end up with an exposed area of $2Rh$

It's easy to prove there are better approaches. Ie: just doing a single cut of length $d$ like so:

Will result in an exposed area of $dh$. With $N=2$ that'll equal $2Rh$ since $d$ will have to be right at the middle of the circle; but as N increases $d$ and therefore $dh$ will decrease, making it a much better solution. The question is: is it the best solution? Even if it turns out to be for a small value of $h$; if $h$ gets too large relative to $R$ (as mentioned above) then eventually $dh$ will be larger than $\pi R^2$ so it'll make more sense to slice the cake horizontally.

 

[Nik_2213]

Take a slice off the bottom ? The 'exposed' surface becomes the new base, mostly sealing it...

Having watched my beloved wife divide a monolithic cake into layers for 'filling', there are kitchen 'jigs' to do this cleanly. Think 'horizontal cheese-slice'. Single and eg Multi-tier types available...


( Problem related to taking small portions off big chunk of cheese to be eaten with eg breakfast rolls: Trick is to stay ahead of excessive desiccation or edible-mould growth... )

 

[adimare]

I appreciate the response. But I'm not really looking to cut a cake; I'm only interested in the mathematical problem. Yes, cutting a wedge from the middle and putting the two halves together would solve the problem in real life (as stated in the OP), same with your solution of cutting a slice from the bottom. However, what I'm looking for is the ideal shape that needs to be cut in order to expose the minimum amount of surface area. I'm pretty sure that when $h$ is very large compared to $R$, then your solution is the best anyways. However, in that case I'd say the exposed area is $\pi R^2$ at the bottom, not 0.