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
[Math] optimization volume of the largest rectangular box
calculusoptimization
calculusoptimization
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
Best Answer
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 $max(V) = 6*2*1 = 12$.