Finding $3$ unknown variables A rectangular prism without a base, with volume $64$, find the length,breadth and height, with the least possible total surface area.
I have established that $V=xyz $ and $\text{SA} = 2xy + xz +2yz$
Is the next step to differentiate the $SA$ and get a minimum value for $x,y,z$?
 A: Apply AM-GM inequality we have: $SA \ge 3\sqrt[3]{4(xyz)^2} = 48\sqrt[3]{4}$, and the equality occurs when $x = z = 2y\implies 4y^3 = 64 \implies y^3 = 16 \implies y = 2\sqrt[3]{2}, x = z = 4\sqrt[3]{2}$.
A: Just as  Bye_World commented, this a problem for which Lagrange mutliplers are really well suited (if you did not hear yet about them, do not worry ! It will be soon and you will enjoy them).
You want to minimize $$S= 2xy + xz +2yz \qquad \text{s.t}  \qquad V=x yz\qquad (V\,\text{being given})$$ So, consider the function $$F=2xy + xz +2yz+\lambda (xyz-V)$$ and compute the partial derivatives $$F'_x=\lambda  y z+2 y+z\tag 1$$ $$F'_y=\lambda  x z+2 x+2 z\tag 2$$ $$F'_z=\lambda  x y+x+2 y\tag 3$$ $$F'_\lambda=x y z-V\tag 4$$ Since you look for an extremum, all the partial derivatives will be equal to $0$.
Using $(1)$,we could get $z=-\frac{2 y}{\lambda  y+1}$; using $(3)$,we could get $y=-\frac{x}{\lambda x+2}$. Replacing in $(2)$ and simplifying, we obtain $$F'_y=x (\lambda  x+4)=0\tag 5$$ Since $x\neq 0$, then the solution is $x=-\frac4 \lambda$. So, at this point we have $$x=-\frac4 \lambda\qquad y=-\frac{2}{\lambda }\qquad z=-\frac{4}{\lambda }$$ Replacing in $(4)$, we then end with $$-\frac{32}{\lambda ^3}-V=0\implies \lambda=-\frac{2^{5/3}}{\sqrt[3]{V}}$$ So, finally $$x=\sqrt[3]{2V}\qquad y=\sqrt[3]{\frac v 4}\qquad z=\sqrt[3]{2V}$$ Now, if you want, plug $V=64$.
Another method would consist in extracting $z$ from the contraint$z=\frac V{xy}$ and to look for the minimum of $$G=2xy + xz +2yz=\frac{2 V}{x}+\frac{V}{y}+2 x y$$ Taking partial derivatives $$G'_x=2 y-\frac{2 V}{x^2}=0\implies y=\frac{V}{x^2}$$ $$ G'_y=2 x-\frac{V}{y^2}=2 x-\frac{x^4}{V}=0\implies x=\sqrt[3]{2V}$$ and the remaining results for $y,z$.
