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I am preparing for an exam tuesday morning and I would like to ask you, if someone could please review my solution for the following excercise. I don't have the correct answer so I am unable to check whether it is OK.

Find the extremes of polynomial function $f(x,y,z) = x^2y + y^2z + x - z$

So here is how i solved it:

$$ \begin{align*} f'_x &= 2xy + 1 \hspace{10mm} f'_y = x^2 + 2yz \hspace{10mm}f'_z = y^2 - 1\\ \end{align*} $$ $$ \begin{align*} f''_{xx} = 2y \hspace{10mm} f''_{xy} = 2x \hspace{10mm} f''_{xz} = 2y\\ f''_yy = 2z \hspace{10mm} f''_{yz} = 2y \hspace{10mm} f''_{zz} = 0\\ \end{align*} $$

So now I need to solve this system:

$$ \begin{align*} 2xy + 1 &= 0\\ x^2 + 2yz &= 0\\ y^2 -1 &= 0 \end{align*} $$

So I get the points:

$A = [-\frac{1}{2}, 1, -\frac{1}{8}]$

$B = [\frac{1}{2}, -1, \frac{1}{8}]$

For point A: $$ \begin{align*} f''{xx}(-\frac{1}{2}, 1, -\frac{1}{8}) = 2 \hspace{10mm} f''{xy}(-\frac{1}{2}, 1, -\frac{1}{8}) = -1 \hspace{10mm} f''{xz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\ f''{yy}(-\frac{1}{2}, 1, -\frac{1}{8}) = -\frac{1}{4} \hspace{10mm} f''{yz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 2 \hspace{10mm} f''{zz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\ \end{align*} $$ $$ H = \begin{pmatrix} 2 & -1 & 0 \\ -1 & -\frac{1}{4} & 2 \\ 0 & 2 & 0 \end{pmatrix} $$ Subdeterminants are: $2, -\frac{3}{2}, -8$

Therefore we don't know whether point A is maxima or minima.

For point B: $$ \begin{align*} f''{xx}(-\frac{1}{2}, 1, -\frac{1}{8}) = -2 \hspace{10mm} f''{xy}(-\frac{1}{2}, 1, -\frac{1}{8}) = 1 \hspace{10mm} f''{xz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\ f''{yy}(-\frac{1}{2}, 1, -\frac{1}{8}) = \frac{1}{4} \hspace{10mm} f''{yz}(-\frac{1}{2}, 1, -\frac{1}{8}) = -2 \hspace{10mm} f''{zz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\ \end{align*} $$ $$ H = \begin{pmatrix} -2 & 1 & 0 \\ 1 & \frac{1}{4} & -2 \\ 0 & -2 & 0 \end{pmatrix} $$ Subdeterminants are: $2, -\frac{3}{2}, 8$

Therefore point B is a local maximum.

So please - is it correct? Thanks in advance.

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It is pretty senseless to maximize/minimize a function with $f(x,0,0)=x$, at least in a global context. –  Listing Oct 22 '11 at 21:16

1 Answer 1

up vote 1 down vote accepted

You wrote $f''_{xz} = 2y$, which is incorrect; it should be 0. However in your later calculations you correctly treat the term as zero.

For your second Hessian, the first subdeterminant is -2, not 2. Even if it were 2, H would not be negative-definite: Sylvester's Criterion states that a matrix is positive-definite (and the critical point a local minimum) if all of the subdeterminants are positive, and negative-definite (and the critical point a local maximum) if the subdeterminants alternate sign, starting with negative. Your second H satisfies neither criterion, and isn't singular (since the determinant is non-zero) so the critical point is neither a minimum nor a maximum.

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I've made an error during transcription. Anyway - thanks. So neither one of these points is minimum nor maximum, right? –  Alexandar Živkovič Oct 22 '11 at 21:44
    
Yes, that's right. –  user7530 Oct 22 '11 at 22:01

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