Question

In a beehive, each cell is a regular hexagonal prism, open at one end with a trihedral angle at the other end as in the figure. It is believed that bees form their cells in such a way as to minimize the surface area, thus using the least amount of wax in cell construction. Examination of these cells has shown that the measure of the apex angle \(\theta \) is amazingly consistent. Based on the geometry of the cell, it can be shown that the surface area \(S\)is given by

\(S = 6h - \frac{3}{2}{s^2}cot\theta + \left( {3{s^2}{{\sqrt 3 } \mathord{\left/

{\vphantom {{\sqrt 3 } 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)csc\theta \)

where, \(s\), the length of the sides of the hexagon, and \(h\), the height, are constants.

Calculate \({{dS} \mathord{\left/

(a) {\vphantom {{dS} {d\theta }}} \right.

(b) \kern-\nulldelimiterspace} {d\theta }}\).

(c)What angle should the bees prefer?

(d)Determine the minimum surface area of the cell (in terms of \(s\)and \(h\)).

Note: actual measurements of the angle \(\theta \) in beehives have been made, and the measures of these angles seldom differ from the calculated value by more than \(2^\circ \).

Short answer

(a)The derivative is\(\frac{{dS}}{{d\theta }} = \frac{3}{2}{s^2}\csc \theta \left( {\csc \theta - \sqrt 3 \cot \theta } \right)\).

(b)The angle the bees should prefer is, \(\theta \approx 54.74^\circ \).

(c) The minimum surface area of the cell is \(6hs + 2.12132{s^2}\).

Step-by-step solution

(a)Step 1: To find the derivative of \(S\) 

We have\(S = 6hs - \frac{3}{2}{s^2}\cot \theta + \left( {3{s^2}{{\sqrt 3 } \mathord{\left/

{\vphantom {{\sqrt 3 } 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)\csc \theta \,\, \cdots \cdots \left( 1 \right)\).

We know theSum Ruleto find the derivative of the sum of the two differentiable functions\(f\left( x \right)\)and\(g\left( x \right)\), which is given by,

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) &= \frac{d}{{dx}}\left( {f\left( x \right)} \right) + \frac{d}{{dx}}\left( {g\left( x \right)} \right)\\ &= f'\left( x \right) + g'\left( x \right)\end{aligned}\)

We know theDifference Ruleto find the derivative of the difference of the two differentiable functions\(f\left( x \right)\)and\(g\left( x \right)\), which is given by,

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right) - g\left( x \right)} \right) &= \frac{d}{{dx}}\left( {f\left( x \right)} \right) - \frac{d}{{dx}}\left( {g\left( x \right)} \right)\\ &= f'\left( x \right) - g'\left( x \right)\end{aligned}\)

Differentiating \(\left( 1 \right)\)with respect to\(\theta \), we get,

\(\begin{aligned}{c}\frac{{dS}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {6hs} \right) - \frac{3}{2}{s^2}\frac{d}{{d\theta }}\left( {{\rm{cot}}\theta } \right) + \left( {3{s^2}{{\sqrt 3 } \mathord{\left/

{\vphantom {{\sqrt 3 } 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)\frac{d}{{d\theta }}\left( {{\rm{csc}}\theta \,} \right)\\\frac{{dS}}{{d\theta }} &= 0 - \frac{3}{2}{s^2}\left( { - {\rm{cs}}{{\rm{c}}^2}\theta } \right) + \left( {3{s^2}{{\sqrt 3 } \mathord{\left/

{\vphantom {{\sqrt 3 } 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)\left( { - {\rm{csc}}\theta \,{\rm{cot}}\theta \,} \right)\\\frac{{dS}}{{d\theta }} &= \frac{3}{2}{s^2}\,{\rm{cs}}{{\rm{c}}^2}\theta - \frac{{3\sqrt 3 {s^2}}}{2}\,{\rm{csc}}\theta \,{\rm{cot}}\theta \\\frac{{dS}}{{d\theta }} &= \frac{3}{2}{s^2}{\rm{csc}}\theta \left( {{\rm{csc}}\theta - \sqrt 3 {\rm{cot}}\theta } \right)\end{aligned}\)

(b)Step 2: To find the angle that bees should prefer

The bees should prefer the angle that minimizes the surface area.

Therefore, we need to find the absolute minimum of the function \(S\left( \theta \right)\).

Thus, the derivative \(\frac{{dS}}{{d\theta }}\) is defined for all \(\theta \) such that

\(\begin{aligned}{c}{\sin ^2}\theta \ne 0\\\sin \theta \ne 0\\\theta \ne \frac{\pi }{2} + k\pi ,\,\,\,\,k \in \mathbb{Z}\end{aligned}\)

Let us find the zeros of the derivative of the function \(S\).

\(\begin{aligned}{l}S'\left( \theta \right) &= 0\\ \Rightarrow 1 - \sqrt 3 \cos \theta &= 0\\ \Rightarrow \sqrt 3 \cos \theta &= 1\end{aligned}\)

\(\begin{aligned}{l} \Rightarrow \cos \theta &= \frac{1}{{\sqrt 3 }}\\ \Rightarrow \theta &= {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)\\ \Rightarrow \theta \approx 54.74^\circ \end{aligned}\)

Here,

\(\begin{aligned}{l}S'\left( \theta \right) > 0,\,\,\,\,\forall \theta > 54.74^\circ \\S'\left( \theta \right) < 0,\,\,\,\,\forall \theta < 54.74^\circ \end{aligned}\)

This implies that the surface area is an absolute minimum at the angle \(\theta \approx 54.74^\circ \).

(c)Step 3: To determine the minimum surface area of the cell

From the part (b), we have seen that the surface function \(S\) has an absolute minimum at \(\theta \approx 54.74^\circ \).

We substitute this value of \(\theta \) in equation \(\left( 1 \right)\)to find the value of surface area.

Therefore, the minimum surface area of the cell is:

\(\begin{aligned}{c}S\left( {54.74^\circ } \right) &= 6hs - \frac{3}{2}{s^2}\cot \left( {54.74^\circ } \right) + \left( {3{s^2}{{\sqrt 3 } \mathord{\left/

{\vphantom {{\sqrt 3 } 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)\csc \left( {54.74^\circ } \right)\\S\left( {54.74^\circ } \right) \approx 6hs - \frac{3}{2}{s^2}\left( {0.70699} \right) + \frac{{3\sqrt 3 {s^2}}}{2}\left( {1.22468} \right)\\S\left( {54.74^\circ } \right) &= 6hs - 1.060485{s^2} + 3.181812{s^2}\\S\left( {54.74^\circ } \right) &= 6hs + 2.12132{s^2}\end{aligned}\)

Hence, the minimum surface area of the cell is \(6hs + 2.12132{s^2}\).