Quick ‘proof’ the taller the can, the more material used:
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?
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.
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.
Quick ‘proof’ the taller the can, the more material used:
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?
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.
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.
Thanks fot the aclaration.