If you optimize a right square pyramid, the ratio of the height and the sidelength of the square will be root 2. Why is this? I feel that it could be because the diagonal of a square with sidelength 1 is root 2 but how does that exactly explain this.
Also its interesting because you'd think an optimal pyramid would try to mimic a sphere or a cube or a cylinder where the ratio of the height and the sidelength/diameter is 1:1, but it doesn't.....
Well, the volume of a square pyramid is given by:
$$\mathcal{V}=\frac{1}{3}\cdot\text{H}\cdot\text{L}^2\tag1$$
Where the base length is given by $\text{L}$ and perpendicular height is given by $\text{H}$.
A right square pyramid with base length $\text{L}$ and perpendicular height $\text{H}$ has surface area of:
$$\mathcal{A}=\text{L}^2+\text{L}\cdot\sqrt{\text{L}^2+\left(2\cdot\text{H}\right)^2}\tag2$$
With a given volume we can solve equation $(1)$ for $\text{H}$:
$$\mathcal{V}=\frac{1}{3}\cdot\text{H}\cdot\text{L}^2\space\Longleftrightarrow\space\text{H}=\frac{\mathcal{V}}{\frac{1}{3}\cdot\text{L}^2}=\frac{3\cdot\mathcal{V}}{\text{L}^2}\tag3$$
Substitute equation $(3)$ into equation $(2)$ gives:
$$\mathcal{A}=\text{L}^2+\text{L}\cdot\sqrt{\text{L}^2+\left(2\cdot\frac{3\cdot\mathcal{V}}{\text{L}^2}\right)^2}\tag4$$
Now, we need to solve:
$$\frac{\partial\mathcal{A}}{\partial\text{L}}=0\tag5$$
And it gives:
$$\text{L}^5\cdot\left(\text{L}+\sqrt{\text{L}^2+\frac{36\cdot\mathcal{V}^2}{\text{L}^4}}\right)-18\cdot\mathcal{V}^2=0\space\Longrightarrow\space\text{L}=\frac{\sqrt[3]{3}\cdot\sqrt[3]{\mathcal{V}}}{\sqrt[6]{2}}\tag6$$
Solving for $\text{H}$, gives:
$$\mathcal{V}=\frac{1}{3}\cdot\text{H}\cdot\text{L}^2\space\Longleftrightarrow\space\text{H}=\sqrt[6]{2}\cdot3^\frac{2}{3}\cdot\mathcal{V}^\frac{2}{3}\tag7$$
So, the fraction is:
$$\frac{\text{H}}{\text{L}}=\frac{\sqrt[6]{2}\cdot3^\frac{2}{3}\cdot\mathcal{V}^\frac{2}{3}}{\left(\frac{\sqrt[3]{3}\cdot\sqrt[3]{\mathcal{V}}}{\sqrt[6]{2}}\right)}=6^\frac{1}{3}\cdot\mathcal{V}^\frac{1}{3}\tag8$$