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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.

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The function obviously has a maximum since it's zero when a variable is zero and is positive when the variables are positive, so if you find a single extreme point, it must be that maximum. – Gerry Myerson Jun 5 '12 at 13:22
up vote 1 down vote accepted

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.

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thank you ! i understood intuitively that the point would have to be a maximum but was wondering about in another case .. I've heard about positive definite matrices a few times, does it say anything about the characteristics of the matrix in the more general sense ? – Ronald Jun 5 '12 at 14:46
Indeed! It is most commonly used for symmetric (hermitian) matrices. For any symmetric matrix there is a basis of eigenvectors and all eigenvalues are real (Spectral theorem). Positive definite means then that all eigenvalues are positive (for negative definite they are all negative). Moreover a positive definite symmetric matrix defines a scalar product, something which behaves like the standard scalar product you know from basic linear algebra. – Simon Markett Jun 5 '12 at 14:54
So another (perhaps more tedious) way of checking postive-definiteness would be computing the eigenvalues and seeing if they're postive .. In a sense could we then think of a positive-definite matrix as the matrix equivalent of a positive real number ? – Ronald Jun 5 '12 at 15:13
The first part is correct. For the second part: I suppose... if you check out the wikipedia article I linked in my answer then you see that it reads a similar statement. However I'd be careful to talk about the equivalent. It certainly has some properties... – Simon Markett Jun 5 '12 at 15:18

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