I find this exercise to be very interesting and irksome that I can't seem to figure it out:
My thought process is as follows:
1.
Our surface area has no lid, so it can be shown as:
$$SA = 2zy + xy + 2xz$$
As there is only $1$ $xy$ square in the surface area. In terms of volume, $x$, and $y$, this is:
$$SA = \frac{2V}{x} + xy + \frac{2V}{y}$$
2.
This part is kind of bizarre to me in what it is exactly asking, but I can make this expression even more in terms of $V$:
$$SA = \frac{2V}{x} + \frac{V}{z} + \frac{2V}{y}$$
Now, I don't see how it's at all illogical to just assume, for the area to be a minimum, we make $x$, $y$, and $z$ tend to $0$ s.t. in our first equation that is only in terms of $x,y,$ and $z$, it will return a very small surface area. However, this is definitely not the answer anyone is looking for.
My lecture proceeded to do the following:
Found the partial derivatives of the expression I found in 1.
Expressed all of them in terms of $V$.
Used the expression $\nabla = F_{xx}F_{yy} - (F_{xy})^2$ to find saddle points and extrema to evaluate whether the expressions made for $x,y,z$ render a minimum, which it did.
However, I don't understand the thought process behind why this answers the question, I guess because I don't understand by "the area to be a minimum".
3.
According to my lecturer, this, in terms of Lagrange multipliers, says the following:
We wish to minimize $2zy + xy + 2xz$ subject to the constraint $V - xyz = 0$. This makes intuitive sense to me. However, I still don't see how there can be a minimized area .
EDIT:
My question seems to boil down to the fact that I haven't yet understood why considering the minimization of the surface area of a box with an open lid is not the trivial let $x,y,z $ be very close but not equal to $0$ which will make the surface area infinitesimally small, and how the constraint $V - xyz = 0$ doesn't allow for this. $x,y,z$ are the side lengths of the box. Winding down the values each close to $0$ will minimze the area, in my view, though the minimizing values $x,y,z$ would not be well defined despite being, in my view, infinitesimally close to $0$.
Well, the volume of the box is given by:
$$\mathscr{V}=x\cdot y\cdot z\tag1$$
And the surface area:
$$\mathscr{S}=y\cdot z+y\cdot z+x\cdot z+x\cdot z+x\cdot y=2\cdot y\cdot z+2\cdot x\cdot z+x\cdot y\tag2$$
Now, we know that the volume should be $\text{V}$, so we can write:
$$x\cdot y\cdot z=\text{V}\space\Longleftrightarrow\space x=\frac{\text{V}}{y\cdot z}\tag3$$
So, we can write for the surface area:
$$\mathscr{S}=2\cdot y\cdot z+2\cdot\frac{\text{V}}{y\cdot z}\cdot z+\frac{\text{V}}{y\cdot z}\cdot y=2\cdot y\cdot z+2\cdot\frac{\text{V}}{y}+\frac{\text{V}}{z}\tag4$$
And now we need to find:
$$ \begin{cases} \frac{\partial\mathscr{S}}{\partial y}=0\\ \\ \frac{\partial\mathscr{S}}{\partial z}=0 \end{cases}\space\space\space\space\space\space\space\space\space\space\space\therefore\space\space\space\space\space\space\space\space\space\space\space y=2^\frac{1}{3}\cdot\text{V}^\frac{1}{3}\space\wedge\space z=\frac{\text{V}^\frac{1}{3}}{2^\frac{2}{3}}\tag5 $$