Isn't the larger the can proportional to how does both top and bottom shrink? like, being the same amount of material, but with a different distribution.

[–] drop_table_username@lemmy.world 3 points 1 day ago (1 children)

No he's right. The solution for an optimal surface area to volume ratio is a sphere. The farther you deviate from a sphere the less optimal you become. The actual math for this is finding deltaSurfaceArea in respects to cylinder radius for a given volume and then finding the maxima, which is a Uni physics 1 problem I really don't feel like doing. Long story short, optimal is when height = diameter, or as close to a sphere as a cylinder can be.

[–] capuccino@lemmy.world 2 points 1 day ago (1 children)

Thanks fot the aclaration.

[–] wuphysics87@lemmy.ml 2 points 14 hours ago (1 children)

It's not really 'right' or 'wrong' it's under a fixed set of assumptions. You raise a valid point. What does happen to the top and the bottom? I was ignoring them considering only the sides in the two most extreme cases.

If I understand your case when the can is flatted the area gets much larger and when it gets taller it shrinks to a pin point. An equally valid approach

[–] capuccino@lemmy.world 1 points 4 hours ago

If I understand your case when the can is flatted the area gets much larger and when it gets taller it shrinks to a pin point.

Yes, that was what I meaning.