Comparison between surface area (wax used) of an optimized regular hexagonal prism and an optimized rhombic dodecahedron (BEE PRISM)

Question

In my other post I was asking about the equation of volume of a rhombic dodecahedron with hexagonal base (the prism which bees use). The area is 3s(2h+(s√2)/2) and the volume is ((3s^2√3)/2)(h-s/(2√2))=0.35 (0.35 is the constraint). Utilizing Lagrange I get sides s and h to be approximately 0.4567 and 0.8073 respectively. This yields a surface area of approximately 2.6547. The thing is that I did it as well with a regular hexagonal prism and the surface area was surprisingly less; approximately 2.254. The area used in that case was 3sa+6sh (becuase the hexagonal aperture is open) and the volume 3ash=0.35. This would mean that bees end up wasting more wax than they should. I thought this might be because in the long run the rhombic dodecahedron might end up tessellating space more efficiently, but I am not sure. Did I go wrong in my operations, or is there a suitable explanation?

Answer 1

Case I: Hexagonal prism

$V = ( 3 \sqrt{3}/2 ) s^2 H = 0.35$

$A = 6 s H + 3 (\sqrt{3}/2) s^2$

Therefore,

$H = 0.7 / (3 \sqrt{3} s^2 )$

Substitute this into $A$

$A = 4.2 / (3 \sqrt{3} s ) + 3 (\sqrt{3}/2) s^2$

Take the derivative with respect to $s$:

$A' = 4.2 / (3 \sqrt{3} ) (-1/s^2) + 3 \sqrt{3} s$

so $A' = 0 $ implies

$s^3 = 4.2 / 27$

Thus,

$s = 0.5378095$

And therefore,

$H = 0.46575673$

and

$A = 2.25439$

Now for the Bee prism

Case II: Bee Prism

Let $H$ be longer height of the lateral sides. Then

Note that $ H = h - \dfrac{s}{\sqrt{2}} $

$V = (3 \sqrt{3}/2 s^2 ) H = 0.35$

To find the surface area of the top part, we need to find the angle it makes with the horizontal plane, and for that, let $v_1$ the vector from the apex (the highest point) along the rhombus side. By a suitable choice of coordinate system with $xy$ plane horizontal, and $x$ axis extending horizontally below the slanted $v_1$, we have

$v_1 = s ( 1, 0, -\dfrac{1}{\sqrt{2}} ) $

Rotating $v_1$ about the $z$ axis by $120^\circ$ counter clockwise, the adjacent edge $v_2$ of the edge $v_1$ is given by

$ v_2 = s ( -\dfrac{1}{\sqrt{2}} , \dfrac{\sqrt{3}}{2} , - \dfrac{1}{\sqrt{2}} ) $

So that the normal to the plane of the rhombus is

$ n = v_1 \times v_2 = ( \dfrac{\sqrt{3}}{2 \sqrt{2}} , \dfrac{3}{2 \sqrt{2}} , \dfrac{\sqrt{3}}{2} ) $

So the angle that the plane of the rhombus makes with the horizontal plane is

$ \cos \psi = \dfrac{ n_z }{\| n \| } = \dfrac{1}{\sqrt{3}} $

From here we can now write the area of whole shape as

$A = 3 (\sqrt{3}/ 2) (\sqrt{3}) s^2 + 6 (1/2) (2 H s - s^2 / \sqrt{2} )$

From the volume expression we have

$H = 0.7 / (3 \sqrt(3) ) (1/s^2)$

Substitute this into the area expression,

$A = 4.5 s^2 + 3 ( 1.4 / (3 sqrt(3) ) (1/s) - s^2 / sqrt(2) ) $

Cleaning up the expression a bit, we get,

$ A = (4.5 - \dfrac{3}{\sqrt{2}} ) s^2 + \dfrac{1.4}{\sqrt{3}} (1/s) $

Differentiate this with respect to $s$,

$A' = ( 9 - \dfrac{6}{\sqrt{2} } ) s + \dfrac{1.4}{\sqrt{3}} (-1/s^2 ) $

Now, $A' = 0 $ implies

$ s^3 = \dfrac{ 1.4 }{ \sqrt{3} ( 9 - \dfrac{6}{\sqrt{2}}) } $

So that,

$ s = 0.55386 $

And this implies that

$ H = 0.439152 $

and

$ A = 2.189 $

Therefore the "BEE prism" consumes slightly less wax for the same enclosed volume.