I used the “bubbles” principle. That is, the maximum volume for any given surface area is spherical, hence the shape of bubbles. I’ll skip the actual physics because it’s a bit hard to explain and my physics is very rusty. But short and fat should be closer to spherical than long and thin. Not a completely robust answer as it ignores the open ends so relies on gut feel.

Master R considered taking things to extremes (a useful technique for problem solving). He said that something extremely long and thin has zero area, as does something extremely short and fat. So something squarish would give a maximum area. Clearly he was simplifying in 2D rather than 3D but his logic seems sound.

I have just worked out why you called the puzzle “Sweets Galore”! I make it about 3,900 smarties!

A quick follow-up teaser. How big (volume) is the fat cylinder relative to the thin one? Can anyone solve it without checking the specific measurements of A4 paper?

The volumes are in the same ratio as the lengths of the sides. If the length of the sides are 2:1 then the volumes likewise are 2:1

Using P = 2.Pi.R sets up the achievable radii. (P = length of paper edge that is curved into a cylinder ie the perimeter of the cylinder)
Using V = Pi.Rsq.H sets up the associated volumes. (H = length of paper edge that forms the height of the cylinder)

Setting up a ratio of V1:V2 results in virtually all the terms cancelling out leaving P/H ie the ratio of the length of the sides.

This what I vizualised when I first saw this teaser. I guess Mike saw something the same or similar ?

Thought I might have a go at setting out my algebra solution to Ravvie’s teaser. More a challenge in the use of PowerPoint actually, so any errors are purely typos

It’s a lot easier to read if you click the enlarge arrows in the bottom right of the picture. You have to hover over the bottom right to find the arrows

I was thinking about the paper teaser last night. If the paper is square, the volumes are the same. If rectangular, we know that the base area is pi x r x r. So the volume is always proportional to the square of the radius, meaning the flatter column will always have the greater volume.

I didn’t know that the A series paper halves, but following the square of the radius, the ratio would be L/W.

(Mike, I know it required knowledge that the A series halves in area each time)

As an aside, A0 has an area of 1m^2, so its sides are √ √2 = 1.189m and 1/√ √ 2 = 0.841m. A4 dimensions are a quarter of that, namely 0.297m by 0.210m. But one only needs to know that it halves in area each time.

I should add that the first measure gave the desired concentration of Rose Feed. Any departure, more concentrated or less concentrated, could cause problems !!

I couldn’t remember the formula for the volume of a cone.

However, if the shape remains the same (as it does here), then doubling each dimension would lead to an 8 times increase in volume, regardless of shape.

Well done again Ravvie. I’m sure that 1/3 Pi R sq H has now firmly re-planted itself in your mind
But, as you neatly pointed out, that info isn’t necessary to get the right solution.

Hi Mike, I did a degree in Civil Engineering at The University of Nottingham way back in the 60’s. Back then, in addition to A Level pure maths, applied maths, physics and chemistry, I had to pass a subject called The Use of English.

First question was along the lines - clearly and concisely describe the process of sewing a replacement button onto the front of a jacket. Describe any tools and materials used.

Apparently, universities were fed up with scientists and engineers who basically couldn’t read or write !