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?? 

Suppose box $B$ with sides $\ell$, $w$, and $h$ has volume $V=\ell wh$, and suppose that $\ell\ne w$. Consider box $B'$, whose sides are $\sqrt{\ell w}$, $\sqrt{\ell w}$, and $h$, which has the same volume.
Box $B$ has surface area $2(\ell w+wh + h\ell)=2\ell w+2h(w+\ell)$, and box $B'$ has surface area $2(\ell w+2h\sqrt{\ell w})=2\ell w+4h\sqrt{\ell w}$. By the arithmetic-geometric mean inequality, $w+\ell\ge 2\sqrt{\ell w}$, and therefore the surface area of $B'$ is smaller.
This observation shows that if a box of volume $V$ has two unequal sides, it does not have the smallest surface area among all boxes of volume $V$. Taking the contrapositive of this implication, If box $B$ does have the smallest surface area of all boxes with volume $B$, then it does not have two unequal sides.
Therefore the box of volume $V$ with the smallest surface area has all sides equal.
