The box has minimum surface area Show that a rectangular prism (box) of given volume has minimum surface area if the box is a cube. 
Could you give me some hints what we are supposed to do?? 
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EDIT: 
Having found that for $z=\frac{V}{xy}$ the function $A_{\star}(x, y)=A(x, y, \frac{V}{xy})$ has its minimum at $(\sqrt[3]{V}, \sqrt[3]{V})$, how do we conclude that the box is a cube?? 
We have that $x=y$. Shouldn't we have $x=y=z$ to have a cube?? 

we will keep the volume at $1.$  let the base have length $x$ and width $y.$  then the volume constraint makes the height of the box $\frac1{xy}.$ you need to minimize the surface area $$A = 2\left(xy+\frac 1x + \frac 1y\right), x > 0, y > 0 $$ now you can use the am-gm inequality $\frac{a+b+c}3\ge (abc)^{1/3}$ to show that $$A \ge 6\left(xy\frac1x \frac 1y\right)^{1/3} = 6.$$ therefore the minimum surface area of the box is $ 6$ subject to the constraint that the volume is $1.$  
if you scale everything, you will get $$A \ge 6V^{2/3}. $$
