You construct a rectangular Box with volume K cm^3 Prove that a cube uses the least amount of material to construct the box
 A: Let $x$, $y$ and $z$ the lenghts of the sides of the box. If we suppose this box is closed, we must to minimize the function $f(x,y,z)=2xy+2xz+2yz$ restricted to $xyz=K$, we can write the volume of the box, $V$, as follows
$$V=2\left(xy+\frac{K}{y}+\frac{K}{x}\right)$$
If we fix $x$ then we have $V_y=2\left(x-\frac{K}{y^2}\right)$ and $V_{yy}=\frac{4K}{y^3}>0$ then $V$ reach its minimun value for this $x$ when $y=\sqrt{\frac{K}{x}}$. Now we are looking where the function $x\mapsto2\left(2\sqrt{Kx}+\frac{K}{x}\right)$ reach its minimum value.
A: Hint: If you know calculus and are allowed to use Lagrange multipliers, use them to find a global maximum for the volume $abc$ assuming the amount of material for the box $2(ab + bc + ca)$ is fixed, and $a,b,c > 0$. It's pretty obvious that a global maximum must exist which is a local maximum for $a,b,c > 0$, and no  local minimum exists for positive $a,b,c$ because you could always make one side arbitrarily small and take the volume to zero.
A: Start from a cube with sides $x$, $y$, and $z$ in length so $x=y=z$. The volume is $xyz$, and the surface area is $2(xy+xz+yz)$. To check if you can change this box so that it has the same volume but smaller area, scale one side by some number $c$ that is greater than 1. That means one of the sides must also scale, but by $1/c$. The surface area is now
$$2(x(cy)+x(\frac{1}{c})+yz)$$
And this is now larger than the previous area because
$$\begin{align*}
x(cy)+x(z/c)&=cxy+\frac{1}{c}xz\\
&=(x^2)(c+\frac{1}{c}),\,\,\,x^2=xy=xz=yz\\
&>x^2
\end{align*}
$$
And so you've now made the area $c+\frac{1}{c}$ times bigger. The same method can be shown for the scaling situation where both $x$ and $z$ scale downward in accordance with $y$ scaling upwards.
A: The AM/GM inequality shows that $2(xy + yz +zx) \ge 6 (xyz)^{\frac{2}{3}}$, with equality only when $x=y=z$.
