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\begin{center}
\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
\end{center}

\textbf{Question}

Find and classify the critical points of the function

$$f(x,y,z)= xyz-x^2-y^2-z^2$$


\textbf{Answer}

For critical points we have
\begin{eqnarray*}
0 = f_1 & = & yz-2x\\
0 = f_2 & = & xz-2y\\
0 = f_3 & = & xy-2z
\end{eqnarray*}

Thus $xyz=2x^2=2y^2=2z^2$, so $x^2=y^2=z^2$.

Hence $x^3=\pm 2x^2$, and $x=\pm 2$ or $0$. Similarly for $y$ and $z$.

The only critical points are $(0,0,0), \ (2,2,2), \ (-2,-2,2), \
(-2,2,-2)$, and $(2,-2,-2)$.

Let $\underline{u}=u\underline{i}+v\underline{j}+w\underline{k}$,
where $u^2+v^2+w^2=1$. Then
\begin{eqnarray*}
D_uf(x,y,z) & = & (yz-2x)u+ (xz-2y)v+ (xy-2z)w\\
D_u(D_uf(x,y,z)) & = & (-2u+zv+yw)u + (zu-2v+xw)v\\
& &  + (yu+xv-2w)w.
\end{eqnarray*}

At $(0,0,0)$, $D_u(D_uf(0,0,0))=-2u^2-2v^2-2w^2<0$ for $\underline{u}
\ne \underline{0}$, so $f$ has a local maximum value at $(0,0,0)$.

At $(2,2,2)$, we have
\begin{eqnarray*}
D_u(D_uf(2,2,2)) & = & (-2u+2v+2w)u + (2u-2v+2w)v\\
& & + (2u+2v-2w)w\\
& = & -2(u^2+v^2+w^2)+ 4(uv+vw+wu)\\
& = & -2[(u-v-w)^2-4vw]
\end{eqnarray*}
$\left \{ \begin{array}{lcl}
<0 & \ \rm{if} \ & v=w=0, \ u\ne 0\\
>0 & \ \rm{if} \ & v=w\ne0, \ u-v-w=0
\end{array} \right. $

Thus $(2,2,2)$ is a saddle point.

At $(2,-2,-2)$ we have:
\begin{eqnarray*}
D_u(D_uf) & = & -2(u^2+v^2+w^2+2uv+2uw-2vw)\\
& = & -2[(u+v+w)^2-4vw]
\end{eqnarray*}
$\left \{ \begin{array}{lcl}
<0 & \ \rm{if} \ & v=w=0, \ u\ne 0\\
>0 & \ \rm{if} \ & v=w\ne0, \ u+v+w=0
\end{array} \right. $

Thus $(2,-2,-2)$ is a saddle point. By symmetry, so are the remaining
two critical points.

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