Minimum amount of material to make a cuboid with fixed volume

Solution 1:

$A(x,y,z)=xy+2xz+2yz ~$ itself has no minimum.

So you should apply the constraint and then take the partial derivatives.

$ \displaystyle A = xy + 2x \cdot \frac{256}{xy} + 2y \cdot \frac{256}{xy} = xy + \frac{512}{y} + \frac{512}{x}$

$ \displaystyle \frac{\partial A}{\partial x} = y - \frac{512}{x^2} = 0$

$ \displaystyle \frac{\partial A}{\partial y} = x - \frac{512}{y^2} = 0$

So, $x^2y = xy^2 = 512 \implies x = y = 8$ and that leads to $z = 4$.