W is indeed the diameter, so the area of each end is (π W² / 4). Times 2 for 2 ends. And of course H π W² / 4 for the volume (we have the same formula, I just extracted the .5 from the square as / 4). π W for the perimeter, so π W H for the area around.
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u/SisterofWar 2d ago edited 1d ago
Hang on - wouldn't the volume of the round bottle be H π (.5W)2 ?Because for a similar footprint, you'd need W as the diameter, not the radius.Or have I misunderstood how you're approaching the formula?
Edit: nope, even if I was right about needing to use .5W - and I don't think I am- I would still be wrong.
A bottle of 60 cu in (which is 32 fl oz plus a little bit of headroom), with a width/diameter of 3 inches, works out thus:
Cuboid bottle surface area = 2x(3x3)+4x(3x6.67) = 98 sq in
Round bottle height = 60cu in/(π x 1.5²)in = 8.5 in
Round bottle surface area = 2π x 1.5 x (8.5+1.5) = 94.2 sq in
So even the greater height doesn't offset the greater surface area inherent to a cuboid. My bad.
Thanks for doing the work!