The point [tex](x,y,z)[/tex] on the plane [tex]x+2y+3z=9[/tex] determines the volume of the box, since [tex]V(x,y,z)=xyz[/tex]. Restricting the box to lie within the first octant is to say that [tex]x,y,z>0[/tex].
Let's do it via Lagrange multipliers. The Lagrangian is
[tex]L(x,y,z,\lambda)=xyz+\lambda(x+2y+3z-9)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=yz+\lambda=0[/tex]
[tex]L_y=xz+2\lambda=0[/tex]
[tex]L_z=xy+3\lambda=0[/tex]
[tex]L_\lambda=x+2y+3z-9=0[/tex]
We have
[tex]L_y-2L_x=xz-2yz=z(x-2y)=0\implies z=0\text{ or }x=2y[/tex]
[tex]L_z-3L_x=xy-3yz=y(x-3z)=0\implies y=0\text{ or }x=3z[/tex]
[tex]2L_z-3L_y=2xy-3xz=x(2y-3z)=0\implies x=0\text{ or }2y=3z[/tex]
We already assume [tex]x,y,z>0[/tex], so we can ignore those options, leaving us with [tex]x=x[/tex], [tex]y=\dfrac x2[/tex], and [tex]z=\dfrac x3[/tex]. Substituting into the plane equation, we get
[tex]x+2\dfrac x2+3\dfrac x3=3x=9\implies x=3\implies y=\dfrac32\text{ and }z=1[/tex]
So the box with largest volume has its vertex (the one opposite the vertex at the origin) in the plane at [tex]\left(3,\dfrac32,1\right)[/tex], giving a volume of [tex]\dfrac92[/tex].
