optimization volume of the largest rectangular box Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes, and one vertex in the plane x +3 y + 6 z = 18
 A: Pick a point $A = (x,y,z)$ on the plane so that $x + 3y + 6z = 18$.
The volume $V$ of the box is $V = xyz$. So we find $max(V)$ with conditions $x + 3y + 6z = 18$ and $x, y, z \in \mathbb{R}^{\text{nonneg}}$.
Use $g(x,y,z) = x + 3y + 6z = 18$. So
$$V'(x) = yz = rg'(x) = r$$
$$V'(y) = xz = rg'(y) = 3r$$
$$V'(z) = xy = rg'(z) = 6r$$
This means $xz = 3yz , xy = 6yz \Rightarrow z(x - 3y) = 0, y(x - 6z) = 0$.
If $z = 0$ or $y = 0$, then $V = 0$ and is not a max. So $$x - 3y = 0 = x - 6z \Rightarrow y = \frac{x}{3}, z = \frac{x}{6}$$
$$\Rightarrow x + 3\left(\frac{x}{3}\right) + 6\left(\frac{x}{6}\right) = 18$$
$$\Rightarrow 3x = 18 \Rightarrow x = 6, y = 2, z = 1$$ So $max(V) = 6*2*1 = 12$.
A: The following is a non-calculus solution. 
We want to maximize $xyz$ subject to the condition $x+3y+6z=18$. 
Let $a=x$, $b=3y$, and $c=6z$. We want to maximize $\frac{abc}{18}$, or equivalently $abc$, subject to $a+b+c=18$.
By the Arithmetic Mean Geometric Mean Inequality, known to "contest kids" as AM/GM, we have for non-negative $a,b,c$ that
$$\frac{a+b+c}{3}\ge (abc)^{1/3},$$
with equality precisely when $a=b=c$. 
Thus $abc$ is maximized when $x=3y=6z$. The maximum value has $abc=(18/3)^3$, and hence $xyz=12$. 
