"The largest cone with the smallest surface area to volume ratio"

That sounds like two things in one.

1) work out which height/radius produces the smallest surface area for volume.
2) work out how to most efficiently place that template on to the 3x3 paper.

BUT it may be possible to make a larger cone from that 3x3 which breaks the "smallest surface area to volume ratio" rule, so which has the priority?

Anyway, you could take your first step by working out which height/radius produces the smallest surface area for volume.

...

The volume of a Cone =  π × r² × (h/3)

The areas of a normal cone are:
Base =  π × r²
Surface Area of Side = π × r × √(r² + h²)

So the ratio would be:

π × r² × (h/3)
---------------------------------
π × r² + π × r × v(r² + h²)

You can cancel π and r easily to produce:

r × (h/3)
---------------------------------
r + √(r² + h²)

I think the net of a cone is a rectangle with a circle stuck on it. The circle will become the base of the cone and the rectangle will wrap around the circle to make the height.

The radius of the circle would be the radius of the base and the length of the rectangle would be the slant height of the cone.

I think MathsIsFun's equation would be easier to work with if it was it terms of slant height instead of height, so let's magically change that:

r x √(s² - r²)
---------------
   3(r + s)

For the rectangle to be able to wrap around the circle, the circle's diameter must not be bigger than 3/π, so the radius must not be bigger than 3/2π.

Also, the slant height and the diameter must not add up to more than 3.

So, basically, we want the highest possible value of [r x √(s² - r²)]/3{r+s}, given the constraints that r ≤ 3/2π and that r + s ≤ 3.

MathsIsFun should use the colon-codes rather than copy-pasting for symbols.

I suppose that makes the majority of my last post obsolete.

This development makes working it out much harder, and so more fun.