Finding maximum volume A box has corner (0,0,0) and all edges parallel to the axes. If the opposite corner (x,y,z) is on the plane $$ 3x+2y+z=1 $$, what position gives maximum volume? Show first that the problem maximizes 
$$
xy-3x^2y-2xy^2
$$
Can somebody please explain how this should be solved? 
 A: From the description of the problem, we have each edge parallel to an axis, one corner at $(0,0,0)$ and one corner at $(x,y,z)$. This means that we can think about the sides of the box each having length $x$, $y$, and $z$. This gives us an area function $A(x,y,z)=xyz$.
We know that the corner $(x,y,z)$ is on the plane defined by $3x+2y+z=1$. If we solve this equation for $z$, we have $z=1-3x-2y$. We can the substitute this into our area function and have $$A(x,y,z)=xy(1-3x-2y)=xy-3x^2y-2xy^2.$$
Now, we need to maximize this function. To get started, we need to take partial derivatives of $A$ with respect to both $x$ and $y$. Please ask if you need help with this part.
A: You want to maximize $|xyz|$ w.r.t. constraint $3x+2y+z=1.$  The hint given in the problem is eliminating variable $z.$
A: Are you familiar with the arithmetic mean/geometric mean inequality? This immediately gives $ 3x = 2y = z $ as the unique absolute maximum of $ xyz $ over positive values of x, y and z, with the given constraint.
