How do I properly set up this optimization equation?

Question

So I've been the given the task to fully optimize any packaging. I chose a DS game box. So first I took the measurements of the cartridge itself ($3.5 \text{ cm} \times 3.3 \text{ cm} \times 0.38 \text{ cm}$). The volume of this game is $4.3\text{ cm}^3$.

So instead of using three variables I decided to make the length and the width ratios of each other. $\frac{3.5\text{ cm}}{3.3\text{ cm}}=\frac{35}{33}$

So...

Constraint: $4.3=\frac{35}{33}x \times (x) \times (y)$

Where $y= \frac{4.3}{\frac{35}{33}x^2}$

Then my optimization equation for the surface area was $SA=2\big(\frac{35}{33x}(x)\big) + 2\big(\frac{35}{33x}(y)\big) + 2\big(xy\big)$

I then did what I normally do for an optimization question where I expanded and then took the derivative of that equation. I then set that derivative to $0$ and solved for $x$, however the answer I received was completely unreasonable.

So disregarding the practicality of optimizing the packaging of a DS game to shrink fit the game, where did I go wrong?

Answer 1

If I understood what you want to do, the formulation of your question will be that you want to minimize ( or maximize) the surface of a cuboid up to a fixed volume :

let me tell you that in practice your problem have to add some constraints about the dimension (the geometry) of what you want put it in.

your objective function is : $$f(x,y,z)= 2(xy+xz+yz)$$ and your constraint is $$xyz=\alpha$$ ( $\alpha$ is a fixed positive number, for your problem $\alpha= 4.3$ )

if you want to maximize, your problem will be meaningless, because, for example, when x became very small y and z in some way have to be very large to hold the volume fix and your f reach infinity.

if you want to minimize you can consider in the first time, the function $f$ with fixed $x$ $(x=x_0)$ : and minimize the function $$f(y)=x_0y+\frac{\alpha}{x_0}+\frac{\alpha}{y}$$ The study of this function will give that we have a minimum at the point $y*=\sqrt{\frac{\alpha}{x_0}}$ after that you try to find the minimum of the function $$x_0 \longmapsto f(y*)$$

you will find that we have a minimum at the point $x_0=\alpha^{\frac{1}{3}}$

so you find out that : $$x=y=z=\alpha^{\frac{1}{3}}$$