[Math] Finding the dimensions of a cuboid for minimal surface area

Question

I have no idea how to even start thinking with this problem:

Using the theorem for extrema of a function with two variables, find the dimensions of a parallelepiped with rectangular faces and fixed volume V such that its surface area is minimal.

Answer 1

Let $P$ be a parallelepiped of dimensions $a,b,c$, its volume is given by $V=abc$ and its surface is $S=2ab+2bc+2ac$. Let us assume that $V>0$, then $c=\frac{V}{ab}$ and thus $S=2ab+2\frac{V}{a}+2\frac{V}{b}$. Now, we look for a maximum of $(a,b)\mapsto S=S(a,b)$ on $\{(x,y)\mid x,y>0\}$. Note that every maximum of $S$ is a critical point $(a^*,b^*)$ of $S$ and satisfy $\nabla S(a^*,b^*)=0$.

Can you take it from here?

The solution is $a^*=b^*=\sqrt[3]{V}$ so that the surface is maximized when the parallelepiped is a cube.