Extreme values of a function on a set

Question

I have a function $\ f(x,y,z)=xyz$ on a set $\ M=\{x,y,z:x+y+z=3\} $ and have to find extreme values of the function on set $M$. I made Lagrange's function $$ L(x,y,z,\Lambda) =xyz+\Lambda x+\Lambda y+\Lambda z-3\Lambda$$ Then I took partial derivatives and counted points where they were all equal to zero. I got four points: $$[1,1,1] ; [3,0,0] ; [0,3,0] ; [0,0,3]$$ To find out whether there is minimum and maximum I counted Hessian matrix: $$\begin{pmatrix} 1 & z & y \\ z & 1 & x \\ y & x & 1 \\ \end{pmatrix} $$ which is indefinite all points but $[1,1,1]$ in which it's not positive, negative definite or indefinite.

How do I know whether there is an extreme value in this point ?

Are you assuming non-negative variables? – user251257 Mar 07 '16 at 20:01 — user251257, Mar 07 '16 at 20:01
No, I am solving it for the whole plane x+y+z=3 – ratrt13 Mar 07 '16 at 20:16 — ratrt13, Mar 07 '16 at 20:16
Then, the problem is unbounded – user251257 Mar 07 '16 at 20:20 — user251257, Mar 07 '16 at 20:20
And how does that help/complicate the solution ? – ratrt13 Mar 07 '16 at 20:31 — ratrt13, Mar 07 '16 at 20:31

score 1 · Accepted Answer · answered Mar 07 '16 at 22:24

$x=y=z=1$ is a local maximizer.

Notice that we can write each feasible point as $(x,y,z) = (1+s, 1-s+t, 1-t)$. It follows $$ 1 - f(x,y,z) = 1- (1+s)(1-s+t)(1-t) = s^2 - st - s^2t + t^2 + st^2 > 0 $$ if and only if $$ s^2 + t^2 > st(1+s-t). $$ For $|s|, |t|$ sufficiently small, the right hand side is around $st$. But, we have $s^2+t^2 \ge 2st$ for any $s,t$. So $(1,1,1)$ is a local maximizer.

Extreme values of a function on a set

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