Is a regular tetrahedron the most optimal equilateral triangle based pyramid and if so how would you prove it?

Question

If one is to find the minimum surface area to enclose a given volume for this type of pyramid, would they arrive at a tetrahedron?

Answer 1

Let $T$ be the base pyramid, and assume the area of $T$ is $B$. Assume the midpoint of $T$ is the origin. Let $P = (x,y,z)$ be the fourth point of the pyramid. Then the volume of the pyramid is $\frac{z \cdot B}{3}$. So planes of the form $z = C$ describe pyramids with equal volume. We now try to find the optimal position of $P$ in such a plane.

If the area of $T$ is $B$, then its base points can be chosen as $(\alpha, 0, 0)$, $(-\frac{\alpha}{2}, -\frac{\alpha \sqrt{3}}{2}, 0)$ and $(-\frac{\alpha}{2}, \frac{\alpha \sqrt{3}}{2}, 0)$ where $\alpha = \frac{2 \sqrt{B}}{\sqrt{3 \sqrt{3}}}.$ It's up to you to calculate the area of the pyramid in function of $x$ and $y$, for a given $z$ and $B$. The desired result is that $(x,y) = 0$. This implies that the optimal pyramids are those whose fourth point lies straight above the midpoint; proving that the tetrahedron is optimal can then be done considering only those kind of pyramids.