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.