Maximize the volume of a rectangular box in the first octant with one vertex in the plane $x+2y+3z=3$ Find the volume of the largest rectangular box in the first octant with the three faces in the coordinate planes and one vertex in the plane $x+2y+3z=3$.
Now I know that $V=xyz$ and I have set $$V=xy\left(1-\frac{x}{3}-\frac{2y}{3}\right).$$
After I take the partial derivatives with respect to $x$ and $y$, I am not sure what to do. Any help would be great! Thanks.

Since $V=xy-\frac{1}{3}x^2y-\frac{2}{3}xy^2$, 
$V_{x}=y-\frac{2}{3}xy-\frac{2}{3}y^2=0$ and $V_{y}=x-\frac{1}{3}x^2-\frac{4}{3}xy=0$.
Multiplying by 3 in each equation and dividing by $y$ in the first equation and by $x$ in the second equation (since $x>0$ and $y>0$) gives
$2x+2y=3$ and $x+4y=3$.  (Now solve for x and y.)
