^{o}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 calculated that the surface area

*S*

_{0}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

*S*

_{1}of this reformed prism no longer the same as

*S*

_{0}. It is not difficult to show that the

*S*

_{1}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 Δ

*S*. This is achieved by setting the first derivative of Δ

*S*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 70

^{o}32'.

Nature is a stingy mathematician.

## 2 comments:

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!

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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