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?? 
$$$$ 
EDIT: 
Having found that for $z=\frac{V}{xy}$ the function $A_{\star}(x, y)=A(x, y, \frac{V}{xy})$ has its minimum at $(\sqrt[3]{V}, \sqrt[3]{V})$, how do we conclude that the box is a cube?? 
We have that $x=y$. Shouldn't we have $x=y=z$ to have a cube?? 
 A: Hint: Let a box be $x$-by-$y$-by-$z$. Here we assume $x>0$, $y>0$, $z>0$. Then the surface area of the box is 
$$A=2(xy+xz+yz)$$
and the volume $V=xyz$ is fixed. We need to find the minimum of $A(x,y,z)$ given additional condition $V=xyz$. But $z=\frac{V}{xy}$ and, hence, $A$ can be considered as a function in two variables $x$ and $y$:
$$A_\star(x,y)=2\left(xy+\frac{(x+y)V}{xy}\right).$$
You need to find the minimum of $A_\star(x,y)$. 
So let 
$$
  \left\{
   \begin{array}{l}
    \frac{\partial A_\star}{\partial x} = 0,\\
    \frac{\partial A_\star}{\partial y} =0;
   \end{array}
  \right.
$$
and so on ... The minimum will be at $x=y=\sqrt[3]{V}$.
Update:
$$
  \left\{
   \begin{array}{l}
    \frac{\partial A_\star}{\partial x} = \frac{\partial}{\partial x} 
    \left( 2xy +\frac{2V}{x}+\frac{2V}{y}\right)=2y-\frac{2V}{x^2}=
    \frac{2x^2y-2V}{x^2}=0,\\
    \frac{\partial A_\star}{\partial y} = \frac{\partial}{\partial y} 
    \left( 2xy +\frac{2V}{x}+\frac{2V}{y}\right)=2x-\frac{2V}{y^2}=
    \frac{2xy^2-2V}{y^2}=0.
   \end{array}
  \right.
$$
Since $x > 0$ and $y > 0$ we obtain
$$
  \left\{
   \begin{array}{l}
    x^2y-V=0,\\
    xy^2-V=0;
   \end{array}
  \right.
  \qquad\text{or}\qquad
  \left\{
   \begin{array}{l}
    x=\sqrt[3]{V},\\
    y=\sqrt[3]{V};
   \end{array}
  \right.
$$
Now we'll proof that $(x_0,y_0)=(\sqrt[3]{V},\sqrt[3]{V})$ is the minimum of $A_\star(x,y)$. Consider
$$
  \begin{bmatrix}
   \frac{\partial^2 A_\star}{\partial x^2} & \frac{\partial^2 A_\star}{\partial x \partial y} \\
   \frac{\partial^2 A_\star}{\partial x \partial y} & \frac{\partial^2 A_\star}{\partial y^2} 
  \end{bmatrix}_{(x_0,y_0)}
=
  \begin{bmatrix}
   \frac{4V}{x^3} & 2 \\
   2 & \frac{4V}{y^3} 
  \end{bmatrix}_{(x_0,y_0)}
=
  \begin{bmatrix}
   4 & 2 \\
   2 & 4 
  \end{bmatrix}
$$
We have the minimum if the matrix above is positively defined. Or, in other words, using Sylvester's criterion we should obtain 
$$
  \left\{
   \begin{array}{l}
    \left.\frac{\partial A_\star^2}{\partial x^2}\right|_{(x_0,y_0)} >0,\\
    \left.\frac{\partial A_\star^2}{\partial x^2}\right|_{(x_0,y_0)}
    \left.\frac{\partial A_\star^2}{\partial y^2}\right|_{(x_0,y_0)}-
    \left( \left.\frac{\partial A_\star^2}
    {\partial x \partial y}\right|_{(x_0,y_0)} \right)^2 >0.
   \end{array}
  \right.
$$ 
Obviously, it is the case. So, $(x_0,y_0)=(\sqrt[3]{V},\sqrt[3]{V})$ is the minimum. Finally, $z_0=\frac{V}{x_0y_0}=\sqrt[3]{V}$. Hence, $x_0=y_0=z_0=\sqrt[3]{V}$ gives the minimum of the area. Consequently, the box should be a cube.
A: Using the method of Lagrange Multipliers, we wish to minimize $$f(x,y,z)=2(xy+xz+yz)$$ subject to the constraint that $$g(x,y,z)=xyz=V$$ for some constant $V$. Differentiating and introducing our Lagrange Multiplier $\lambda$, $$\frac{\partial f}{\partial x}=\lambda \frac{\partial g}{\partial x}$$ $$ \frac{\partial f}{\partial y}=\lambda \frac{\partial g}{\partial y} $$ $$ \frac{\partial f}{\partial z}=\lambda \frac{\partial g}{\partial z} $$ we find that $$2(y+z)=\lambda yz$$ $$2(x+z)=\lambda xz$$ $$2(x+y)=\lambda yx$$  and that therefore $$\frac{x+y}{xy}=\frac{x+z}{xz}=\frac{y+z}{yz}$$ From which we can quickly discover that $x=y=z$. Beating the dead horse, we plug back into our constraint equation to find that $x=y=z=\sqrt[3]{V}$, as expected. 
A: we will keep the volume at $1.$  let the base have length $x$ and width $y.$  then the volume constraint makes the height of the box $\frac1{xy}.$ you need to minimize the surface area $$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}. $$
A: Suppose box $B$ with sides $\ell$, $w$, and $h$ has volume $V=\ell wh$, and suppose that $\ell\ne w$. Consider box $B'$, whose sides are $\sqrt{\ell w}$, $\sqrt{\ell w}$, and $h$, which has the same volume.
Box $B$ has surface area $2(\ell w+wh + h\ell)=2\ell w+2h(w+\ell)$, and box $B'$ has surface area $2(\ell w+2h\sqrt{\ell w})=2\ell w+4h\sqrt{\ell w}$. By the arithmetic-geometric mean inequality, $w+\ell\ge 2\sqrt{\ell w}$, and therefore the surface area of $B'$ is smaller.
This observation shows that if a box of volume $V$ has two unequal sides, it does not have the smallest surface area among all boxes of volume $V$. Taking the contrapositive of this implication, If box $B$ does have the smallest surface area of all boxes with volume $B$, then it does not have two unequal sides.
Therefore the box of volume $V$ with the smallest surface area has all sides equal.
A: Volume $V=xyz$ given. Area $A(x,y,z)=2(xy+yz+zx)$, to minimise, when $x,y,z>0$ and $xyz=V$.
Fact. If $a,b,c>0$, then $a+b+c\ge 3\sqrt[3]{abc}$, and equality holds if and only if $a=b=c$. 
Proof. We set $X=\sqrt[3]{a}$, $Y=\sqrt[3]{b}$ and $Z=\sqrt[3]{c}$. Then the identity
$$
X^3+Y^3+Z^3-3XYZ=\frac{1}{2}(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big),\tag{1}
$$
holds. This means that 
$$
a+b+c-3\sqrt[3]{abc}=X^3+Y^3+Z^3-3XYZ\ge 0,
$$
as the right-hand-side of $(1)$ is non-negative, and the equality only if $X-Y=Y-Z=Z-X=0$ or $X=Y=Z=0$ or $a=b=c=0$. $\quad\Box$
Hence,
$$
\frac{A}{2}=xy+yz+zx\ge 3\sqrt[3]{xy\cdot yz\cdot zx}=3\sqrt[3]{x^2y^2z^2}=3V^{2/3}.
$$
and equality holds iff $xy=zx=yz$ or equivalently iff $x=y=z$.
Indeed, $A$ is minimised when $x=y=z$, and $A_{\mathrm{min}}=6V^{2/3}$.
A: Yes you would have that $x=y=z$, but you have that $V=xyz$ for a rectangular prism, as we have that $x=y=V^{1/3}$ this gives us that $V=V^{2/3}z$ which implies that $z=V^{1/3}$ which means that $x=y=z$
A: What part is not clear? The symmetry automatically pulls you into other two situations cyclically.
Let us take Lagrange Multiplier (as others have also done).
I take the unified Lagrangian combining object and constraint functions of 
Volume and Area together. 
I also choose it such that in $ V - A \cdot \lambda \tag{1}$ $\lambda$  would be
physically a  linear dimension for a side of a rectangular parallelepiped,except 
for a constant factor.
$ x y z - ( x y + y z + z x) \lambda \tag{2}$
Partial differentiation with respect to x gives $ y z - ( y+z) \lambda =0,\, \dfrac1y + \dfrac1z= \dfrac{1}{\lambda} \tag{3}$
Remember that when number of independent variables are more than 2, partial differentiation should be done with respect to each variable.
So similarly by cyclic symmetry, $ z x -( z + x)\lambda =0 , \,\dfrac1z + \dfrac1x= \dfrac{1}{\lambda} \tag{4} $
and
$ xy -( x+y)\lambda =0 , \,\dfrac1x + \dfrac1y= \dfrac{1}{\lambda} \tag{5} $
Summming up the three and halving,
$ \dfrac1x + \dfrac1y + \dfrac1z= \dfrac{3}{2\lambda} \tag{6} $
Subtracting from this the second part of $  (3), (4), (5)$ we get
$ \dfrac1x = \dfrac{1}{2 \lambda} \tag{7}, x = 2 \lambda $
that gives you
$ x = y = z = 2 \lambda = a , $ say.
So finally $ V = a^3$ and $ A = 6 a^2.$
