Consider two cases ignoring the top and bottom only focussing on the surface area. In the first case, you flatten so much the can has no height. This forms a ring that when unwrapped makes a length of 2 pi R.

Now stretch the can to be 'infinitely' long. By construction, this is longer than 2 pi r. Given both are made of aluminum, and have the same density, the larger can has more mass requiring more material.

The total mass must be a continuous function ranging from the linear mass density times the circumference of the circle to the same mass density time times the 'length' of the infinite line. This must remain true for any small increase in length between the two.

I'll leave this as an exercise to the reader. What if the circle has an infinite radius?

[–] capuccino@lemmy.world 2 points 22 hours ago (1 children)

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 21 hours 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 21 hours ago (1 children)

Thanks fot the aclaration.

[–] wuphysics87@lemmy.ml 1 points 10 hours ago

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