Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.