(Solved): Show that a rectangular box of given volume and minimum surface area is a cube. Let x,y, and z ...

Show that a rectangular box of given volume and minimum surface area is a cube. Let $x,y$, and $z$ be the length, width, and height, respectively, and let $v_0$ be the given volume. Then $v_0=xyz$ and $z=\frac{v_0}{xy}$. Write the surface area as a function of two variables. Now, taking partial derivatives, we have $\begin{array}{l}{s}_x=\\ s_y=\end{array}$ Setting each partial derivative equal to zero, we have $\begin{array}{l}{S}_x=?v_0=0\\ s_y=?v_0=0\end{array}$ Solving simultaneously yields $x=$ $y=$ $z=$