Question

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 + 2y + 3z = 9

Answer 1

The point $(x,y,z)$ on the plane $x+2y+3z=9$ determines the volume of the box, since $V(x,y,z)=xyz$. Restricting the box to lie within the first octant is to say that $x,y,z>0$.

Let's do it via Lagrange multipliers. The Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x+2y+3z-9)$

with partial derivatives (set equal to 0)

$L_x=yz+\lambda=0$
$L_y=xz+2\lambda=0$
$L_z=xy+3\lambda=0$
$L_\lambda=x+2y+3z-9=0$

We have

$L_y-2L_x=xz-2yz=z(x-2y)=0\implies z=0\text{ or }x=2y$
$L_z-3L_x=xy-3yz=y(x-3z)=0\implies y=0\text{ or }x=3z$
$2L_z-3L_y=2xy-3xz=x(2y-3z)=0\implies x=0\text{ or }2y=3z$

We already assume $x,y,z>0$, so we can ignore those options, leaving us with $x=x$, $y=\dfrac x2$, and $z=\dfrac x3$. Substituting into the plane equation, we get

$x+2\dfrac x2+3\dfrac x3=3x=9\implies x=3\implies y=\dfrac32\text{ and }z=1$

So the box with largest volume has its vertex (the one opposite the vertex at the origin) in the plane at $\left(3,\dfrac32,1\right)$, giving a volume of $\dfrac92$.