The Mathematics of Honeycomb: Part II

As explained in Part I of the article, instead of building their house the easier way with hexagonal top and bottom, the honeybees are building their cells with hexagonal top and 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 70o32'.

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 calculated that the surface area S0 and the volume V of this prism are:

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 S1 of this reformed prism no longer the same as S0. It is not difficult to show that the S1 is

The difference between the two surface areas (i.e. the amount of wax saved) is thus:

To minimize the amount of wax used, the honeybees are interested in maximizing the value of ΔS. This is achieved by setting the first derivative of ΔS to zero:

The solution to this equation is: x = a/√8. If you plug this value into the honeycomb cell geometry, you will understand why the honeybees choose the rhombic angle to be 70o32'.

Nature is a stingy mathematician.


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
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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