Pyramids construction

We consider the height $h$ of the pyramid as a varyable parameter, whereas its side length $a$ is fixed by the volume $V = \frac{1}{3} a^2 h = \text{const} \quad \Rightarrow \quad a = \sqrt{\frac{3 V}{h} }$ Pythagorean theorem yields the height $h'$ of the triangular area $h' = \sqrt{\frac{a^2}{4} + h^2} = \sqrt{\frac{3 V}{4 h} + h^2}$ so that the total outer area (4 triangles) results to $A = 4 \cdot \frac{1}{2} a h' = 2 \sqrt{3 V} \sqrt{\frac{3 V}{4 h^2} + h}$ In the minimum, the derivative must be zero $\begin{aligned} & & \frac{\partial A}{\partial h} &= \sqrt{3 V} \frac{- \frac{3 V}{2 h^3} + 1}{\sqrt{\frac{3 V}{4 h^2} + h} } \stackrel{!}{=} 0 \\ \Rightarrow & & h^3 &= \frac{3}{2} V = \frac{1}{2} a^2 h \\ \Rightarrow & & h &= \frac{1}{\sqrt{2}} a \\ \Rightarrow & & \theta &= \arctan \frac{2h}{a} = \arctan \sqrt{2} \approx 54.7356 \end{aligned}$

Let $b = \tfrac12a$ be half of the length of a side. The volume is proportional to $V \propto b^2h.$ The area is proportional to $A \propto bh' \propto b\sqrt{b^2+h^2};$ in order to minimize the area, we might as well minimize its square, i.e. we minimize $b^4 + b^2h^2.$ We wish to express this polynomial (of fourth degree) in terms of the constant $V \propto b^2h$ (third degree) and the unknown $t = \tan\theta = h/b$ (zeroeth degree). This must be done as follows: $b^4 + b^2h^2 \propto V^{4/3}(t^{-4/3} + t^{2/3}).$ Thus we minimize the expression $t^{-4/3} + t^{2/3}$ . The easiest way to do this is by setting the derivative equal to zero: $\frac{-4}3 t^{-7/3} + \frac 2 3 t^{-1/3} = 0;$ multiply by $\tfrac32 t^{7/3}$ : $-2 + t^2 = 0.$ We find that $\tan\theta = t = \sqrt 2$ , and so $\theta \approx \boxed{54.7356^\circ}$ .

Alternative

Since the volume is constant, we have (with $t = \tan\theta = h/b$ ) $0 = d(b^2h) = 2bh\:db + b^2\:dh = b^2(2t\:db + dh),$ so that $dh = -2t\:db$ .

To minimize the area we minimize $b^4 + b^2h^2$ (see above); we set its total derivative equal to zero: $0 = d(b^4 + b^2h^2) = (4b^3+ 2bh^2)db +2b^2h\:dh = 2b^3\left[(2 + t^2)\:db + t\:dh\right];$ substitute $dh = -2t\:db$ , we see that $0 = (2 + t^2)\:db + t\:dh = (2+t^2- 2t^2)db = (2-t^2)db,$ from which $t = \sqrt 2$ .

I wrote a python program to optimize the volume versus the top surface area using the functions 3V/a^2=h , hprime =a/cos(theta) , and SAtop=2*hprime*a .

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V=input("What is the volume: ")
V=float(V)
a_alow=(3*V)**(1/3)
h_alow=a_alow
import math
theta_alow=math.atan(h_alow/a_alow)
h_alow_prime=a_alow/math.cos(theta_alow)
SA_top_alow=2*h_alow_prime*a_alow
a_ahigh=(3*V+1)**(1/3)
h_ahigh=3*V/a_ahigh**2
theta_ahigh=math.atan(h_ahigh/a_ahigh)
h_ahigh_prime=a_ahigh/math.cos(theta_ahigh)
SA_top_ahigh=2*h_ahigh_prime*a_ahigh
while True:
   if SA_top_ahigh>SA_top_alow:
     a_ahigh=a_alow  
     h_ahigh=h_alow
     theta_ahigh=math.atan(h_ahigh/a_ahigh)
     h_ahigh_prime=a_ahigh/math.cos(theta_ahigh)
     SA_top_ahigh=2*h_ahigh_prime*a_ahigh
     a_alow=a_alow*.99
     h_alow=3*V/a_alow**2
     theta_alow=math.atan(h_alow/a_alow)
     h_alow_prime=a_alow/math.cos(theta_alow)
     SA_top_alow=2*h_alow_prime*a_alow
  else:
     a_alow=a_ahigh
     h_alow=h_ahigh
     theta_alow=math.atan(h_alow/a_alow)
     h_alow_prime=a_alow/math.cos(theta_alow)
     SA_top_alow=2*h_alow_prime*a_alow
     a_ahigh=a_ahigh*1.01
     h_ahigh=3*V/a_ahigh**2
     theta_ahigh=math.atan(h_ahigh/a_ahigh)
     h_ahigh_prime=a_ahigh/math.cos(theta_ahigh)
     SA_top_ahigh=2*h_ahigh_prime*a_ahigh
print("theta alow is: "+str(theta_alow)+", and SA top is: "+str(SA_top_alow))
print("theta ahih is: "+str(theta_ahigh)+", and SA top is: "+str(SA_top_ahigh)) 

My solution was similar to Rocco Dalto's. I'm just amazed at how close to 54 degrees the answer was, but that it wasn't actually 54 degrees. I was initially expecting a relationship to a golden angle.

Using the diagram above:

Volume $V = \dfrac{1}{3}a^2h$ , where $a = 2h' \cos\theta, h = h' \sin\theta \implies$

$V = \dfrac{4}{3} h'^3 \cos^2\theta \sin\theta = K \implies h' = (\dfrac{3K}{4 \cos^2\theta \sin\theta})^\dfrac{1}{3} \implies$

Lateral Surface Area $S = 2 a h' \implies S(\theta) = 4 h'^2 \cos\theta = j (\sec\theta)^\dfrac{1}{3} (\csc\theta)^\dfrac{2}{3}$ , where $j$ is just a constant.

$\dfrac{dS}{d\theta} = \dfrac{j}{3} (sec\theta)^\dfrac{1}{3} (\csc\theta)^\dfrac{2}{3} * (\tan\theta - 2 \cot\theta) = 0 \implies$

$3 \sin^2\theta - 2 = 0 \implies \sin\theta = \pm \sqrt{\dfrac{2}{3}}$ . choosing $\sin\theta = \sqrt{\dfrac{2}{3}} \implies \theta = \boxed{54.7356^\circ}$

Note: The angle $\theta$ is the same for any pyramid whose base is a regular $n-gon$ .

To show that angle $\theta$ is the same for any pyramid whose base is a regular $n-gon$ .

For area of $n - gon$ :

Let $BC = x$ be a side of the $n - gon$ , $AC = AB= r$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{180}{n}$ .

$\dfrac{x}{2} = r \sin(\dfrac{180}{n}) \implies r = \dfrac{x}{2 \sin(\dfrac{180}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) \implies A_{\triangle{ABC}} = \dfrac{1}{4} \cot(\dfrac{180}{n}) x^2 \implies$

$A_{n - gon} = \dfrac{n}{4} \cot(\dfrac{180}{n}) x^2 \implies$ the Volume of the pyramid $V_{p} = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H$

Let $AC = H$ be the height of the pyramid, $BC = h^*$ , and $AB = h'$ be the slant height of the pyramid and $m\angle{CBA} = \theta$

The lateral surface area $S = \dfrac{n}{2} x h'$ .

$h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) = h' \cos(\theta) \implies x = 2 \tan(\dfrac{180}{n}) \cos(\theta) h'$ and $H = h' \sin(\theta)$ and letting $u(n) = \tan(\dfrac{180}{n}) \implies$ $V_{p} = \dfrac{n}{3} u(n) \cos^2(\theta) \sin(\theta) h' = K$ and $S = n * u(n) cos(\theta) h'^2.$

$V_{p} = \dfrac{n}{3} u(n) \cos^2(\theta) \sin(\theta) h' = K \implies h' = (\dfrac{3K}{n * u(n_) cos^2(\theta) sin(\theta)})^{\dfrac{1}{3}} \implies S(\theta) = j(n) * (sec(\theta))^{\dfrac{1}{3}} (csc(\theta))^{\dfrac{2}{3}}$ , where $j(n) = (9 k^2 n * u(n))^{\dfrac{1}{3}}$

$\implies \dfrac{dS}{d\theta} = \dfrac{j(n)}{3} (sec\theta)^\dfrac{1}{3} (\csc\theta)^\dfrac{2}{3} * (\tan\theta - 2 \cot\theta) = 0 \implies$

$3 \sin^2\theta - 2 = 0 \implies \sin\theta = \pm \sqrt{\dfrac{2}{3}}$ . choosing $\sin\theta = \sqrt{\dfrac{2}{3}} \implies \theta = \boxed{54.7356^\circ}$

$\therefore$ The measure of the desired angle $\theta$ is independent of $n$ .