Optimisation of a juice box: finding the least possible surface area that can hold the most volume I have an investigation which requires me to design the dimensions of a juice box (cuboid) which has the least possible surface area that can hold the most volume.
I am not sure as to how I should approach this problem. I know that differential calculus would help, but I am not even sure how to start (I am new to calculus).
Any pointers would be much appreciated.. I just need a starting point. Thanks!
 A: Hint:
The formula for the surface area, $A_s$ is,
$$A_s=2 \cdot (x^2+y^2+z^2)$$
The formula for the volume, $V$, is,
$$V=x \cdot y \cdot z$$
We want the maximum volume given that the surface area is constrained to be equal to some number $S$.  Taking partial derivatives, shown by adding a subscript, we should have,
$$V_x=V_y=V_z=0$$
and
$$0=S-2 \cdot (x^2+y^2+z^2)=F$$
If we define a new function $L(x,y,z)$ to be,
$$L=V+\lambda \cdot F$$
Then it's hopefully clear that, since we demand the first three equalities, and are the fourth follows from the definition of $F$, we have,
$$L_x=L_y=L_z=F=0$$
Four equations, four unknowns, so this is solvable. For the astute reader, this method might lead a minimum. In this situation however, the minimum is trivial and thus one of the possible maxima may be discarded as a viable candidate for the maximum. 
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A: You need to find a function, this function must be related to the parameters of the box, in this case its measures, length, height and width. Then for this role you must find the derivative and using the criteria of the first and second derivative to find the maximum and minimum.
