Maximize volume of a box in first octant So the question is 

Find the volume of the largest rectangular box with 3 faces on the co-ordinate planes and one vertex in the plane $x+2y+3z=6$. 

When I started typing this question, I didn't know what to do but as I got into it I tried some things and now actually have an answer. I'm not 100% sure it's right so all I'm asking now is confirmation of my ideas. 
The first thing I tried is to deal with fewer variables, so I get $x$ in terms of $y$ and $z$:
$$x = 6-2y-3z.$$
This then gives the formula for volume: 
$$V= x\cdot y\cdot z = (6-2y-3z)\cdot y\cdot z  = 6yz -2y^2z - 3z^2.$$ 
Then I get the gradient $\Delta V = \left( \begin{array}{ccc}
6z - 4yz - 3z^2 \\
6y - 2y^2 -6yz \end{array} \right).$ 
Setting $\Delta V = 0$ gives 
$$z(6-4y-3z) = 0 \implies z=0 \ \text{ or } \ (6-4y-3z) = 0,$$
$$y(6-2y-6z) = 0 \implies y=0 \ \text{ or } \ (6-2y-6z) = 0.$$
Now $z=0$, $y=0$ surely can't be maximums, so I'm concerned with $$(6-4y-3z) = 0 \ \text{ and } \ (6-2y-6z) = 0$$
Solving gives me $y = 1$, $z = \frac{2}{3}$.     $\quad(*)$
To then check if it is a maximum, we compute the Hessian which will be
$$\text{Hessian} =   \left( \begin{array}{ccc}
-4z & 6-4y-6z  \\
6-4y-6z & -6y  \end{array} \right),$$
and the determinant of this is then $12$  (using $(*)$).
To the best of my knowledge, because $12>0$, this implies that $(*)$ gives a maximum or minimum ... but now how do I check this is a max and not a min? When I tried reading it up I get terms thrown at me like "positive semi-definite" and I'm not really sure what that means. Anyway, I hope I'm right about my thoughts and any of your thoughts will be appreciated.
 A: As the comment suggests you don't really have to go through the hassle to check whether you have a maximum or a minimum. The function is non-negative and takes on the value $0$. If you have only one critical point it has to be a maximum. 
But to give a complete answer: The usual criterion for functions $\mathbb R\to\mathbb R$ that you have a minimum if the second derivative is positive and a maximum if it is negative translates to the following:
The critical point is a maximum if the Hessian is negative definite and a minimum if it is positive definite. To check whether a (symmetric) matrix is positive definite is relatively easy: All leading principal minors have to be positive. In other words the determinants of all upper left $n\times n$ submatrices have to be positive (here you just have the $1\times 1$ and the $2\times 2$ case). To check whether a matrix is negative definite you have a condition that the signs of the principal minors must be alternating. However it is easier to multiply the matrix by $-1$ and check for positive definiteness.
Note that there are more possibilities than postive or negative definite, so this characterisation is not exhaustive.
In you example the principal minors are $-4z=-\frac 83<0$ and $12>0$ (alternating). For minus the Hessian you have $4z>0$ and $12>0$ (all positive). Thus the Hessian is negative definite and you have a maximum.
