The Mathematics of Honeycomb: Part II

As explained in Part I of the article, instead of building their house with the easy method of hexagonal top + hexagonal bottom, honeybees are building their cells with hexagonal top + rhombic bottom. In this article, I will try to give a calculus-based demonstration on how the honeybees achieve the economization of their wax resources with a rhombic angle of 70°32'.
Suppose we start with a hexagonal prism with flat bottom like the one shown in the figure above. Let \( AB = a\), \(A''A = b\). Then it can be shown that the surface area \(S_0\) and the volume \(V\) of this prism are: $$S_0= 6ab + \frac{3\sqrt{3}}{2}a^2, \quad V = \frac{3\sqrt{3}}{2}a^2b$$ respectively. Mark a point \(B'\) on the prism so that \(B'B'' = x\) and slice a tetrahedron from the prism along the line \(A''C''\) through point \(B'\). Then, flip and rearrange the sliced tetrahedron on top of the prism, as shown below.
If you repeat this procedure for the other two sides of the prism, you will end up with a prism shown below.
Since we are doing slicing and rearrangement of parts on the prism, the volume of this rearranged prism is conserved. However, the surface area \(S_1\) of this reformed prism no longer the same as \(S_0\). It is not difficult to show that the \(S_1\) is: $$S_1 = 6ab - 3ax +3a\sqrt{3} \sqrt{x^2 + \tfrac{1}{4}a^2}$$ The difference between the two surface areas (i.e. the amount of wax saved) is thus: $$\Delta S(x) = S_0 - S_1 = \frac{3\sqrt{3}}{2}a^2 + 3ax - 3\sqrt{3}a\sqrt{x^2 + \tfrac{1}{4}a^2}$$ To minimize the amount of wax used, the honeybees must maximize the value of \(\Delta S\). This is achieved by setting the first derivative of \(\Delta S\) to zero: $$\Delta S'(x) = 3a - 3\sqrt{3}ax \left(x^2 + \tfrac{1}{4}a^2\right)^{-1/2} = 0$$ The solution to this equation is: $$x = \tfrac{1}{\sqrt{8}}a$$ Since \(A'A'' = \frac{1}{\sqrt{8}}a\) and \(A'B = a\), the hypotenuse \(A''B'\) must be \(\frac{3}{\sqrt{8}}a\). For a regular hexagon, we have \(A''C'' = a\sqrt{3}\), and therefore the rhombic angle \(GA''B'\) must be $$2 \cos^{-1}\sqrt{\tfrac{2}{3}} = 70.5288 = 70^\circ 32'$$ To understand why Nature is a stingy mathematician. See Part III of this blog series.

Comments

Anonymous said…
Can you please Email me the work showing how x=a/(8^1/2) is substituted into honeycomb cell geometry to arrive at the rhombic angles.
Email me at jdelatorre@dusd.net
Thank you!
Ang Wee Lee said…
Since A'A'' = a/8^0.5 and A'B = a, hypotenuse A''B' is 3a/8^0.5. For a regular hexagon, we have A''C" = a*3^0.5. Now the angle GA"B' be written in term of cosine as 2*acos((2/3)^0.5) = 70.529 degrees

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