Finding the dimensions of a cuboid for minimal surface area 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.
 A: 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.

A: The AM-GM inequality gives $S\ge 6V^{\frac23}$, with equality only when $a=b=c$.
A: If we eliminate z from volume and plug it into area we get half area
$ x y + V ( \frac{1}{x} + \frac{1}{y} ) $
It is possible to take Volume itself as Lagrange Multiplier for defined functions
$ F,G $ as:
$$ {x\cdot y ,  ( \frac{1}{x} + \frac{1}{y} )} = { F,G} $$ differentiate w.r.t. both variables,
$$ \frac{F_x}{F_y} = \frac{G_x}{G_y} $$ which on simplification gives $ x = y $
When y is eliminated $ z = x $ and so we get $ x = y = z = \sqrt[3]{V.}$
