Extension to the Honeycomb Conjecture

The honeycomb conjecture assumes you use equal area shapes.
If you are allowed to use multiple different regular polygons, as long as the length of the side is the same, what pattern would produce the maximum honey to wax ratio?

I am finding this ratio for regular polygon tilings:

The way I am doing this is by finding the smallest area that can be tiled, including both the honey and the wax.

Starting with the honeycomb conjecture's hexagons..

Hexagon Tile

Each hexagon is bordered by 6 rectangles and 6 equilateral triangles.

If I want to tile the hexagon's border, I only need to include 3 rectangles and 2 triangles in each copy:

The area for a hexagon is ${\displaystyle \frac{3\sqrt{3}}{2}{s}^2}$
The area for a rectangle is ${\displaystyle s*w}$
The area for an equilateral triangle is ${\displaystyle \frac{\sqrt{3}}{4}{w}^2}$
Where s is the length of the side of the hexagon, w is the width of the border.

If we make the the length of s variable, while setting w to a constant 1, this ratio equals: ${\displaystyle \frac{3\sqrt{3}{s}^2}{6s+\sqrt{3}}}$

Looks good? Okay, now let's switch to a more complicated tile.

Truncated hexagonal tiling

See Truncated hexagonal tiling on wikipedia

Honey would be inside the dodecagon and inside the surrounding 6 equilateral triangles.

Between each dodecagon and triangle is the wax border.

Notice the borders are different sizes. If we set the thinner borders to 1, then the thicker border between the dodecagons is $\sqrt{3}$.

The area for a dodecagon is ${\displaystyle (6+3\sqrt{3})s^2}$
The area for an equilateral triangle is ${\displaystyle \frac{\sqrt{3}}{4}{s}^2}$
The area for the thin rectangle is ${\displaystyle s*w}$
The area for the thick rectangle is ${\displaystyle \sqrt{3}*s*w}$
Small triangle has angles 30°, 30°, 120°, which actually has the same area as an equilateral triangle: ${\displaystyle \frac{\sqrt{3}}{4}{w}^2}$
Where s is the length of the side of the dodecagon and triangle, w is the width of the border.

If we make the the length of s variable, while setting w to a constant 1, this ratio equals: ${\displaystyle \frac{(12+7\sqrt{3})s^2}{6(2+\sqrt{3})s+3\sqrt{3}}}$

Comparing the two tiles

With hexagons, our honey to wax ratio is ${\displaystyle \frac{3\sqrt{3}{s}^2}{6s+\sqrt{3}}}$

With the truncated hexagonal tiling, the ratio is ${\displaystyle \frac{(12+7\sqrt{3})s^2}{6(2+\sqrt{3})s+3\sqrt{3}}}$

If I leave out the side triangles of the wax, the ratio for the hexagon is ${\displaystyle \frac{\sqrt{3}}{2}s\approx 0.8660254s}$

If I leave out the side triangles of the wax, the ratio for the truncated hexagonal tiling is ${\displaystyle (\frac{1}{2}+\frac{1}{\sqrt{3}})s\approx 1.07735s}$

The honeycomb conjecture assumes you use equal area shapes.
If you are allowed to use multiple different regular polygons, as long as the length of the side is the same, what pattern would produce the maximum honey to wax ratio?