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.