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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
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\textbf{Question}

Find the minimum and maximum values of $$f(x,y,z)=
xyze^{-x^2-y^2-z^2}.$$
How do you know that extreme values exist?


\textbf{Answer}

\begin{eqnarray*}
f(x,y,z) & = & xyze^{-x^2-y^2-z^2}\\
f_1(x,y,z) & = & yz(1-2x^2)e^{-x^2-y^2-z^2}\\
f_2(x,y,z) & = & xz(1-2y^2)e^{-x^2-y^2-z^2}\\
f_3(x,y,z) & = & xy(1-2z^2)e^{-x^2-y^2-z^2}
\end{eqnarray*}
Any critical point must satisfy
\begin{eqnarray*}
yz(1-2x^2) & = & 0\\
& & \rm{i.e.} \ y=0 \ \rm{or} \ z=0 \ \rm{or} \
x=\pm\frac{1}{\sqrt{2}}\\
xz(1-2y^2) & = & 0\\
& & \rm{i.e.} \ x=0 \ \rm{or} \ z=0 \ \rm{or} \
y=\pm\frac{1}{\sqrt{2}}\\
xy(1-2z^2) & = & 0\\
& & \rm{i.e.} \ y=0 \ \rm{or} \ y=0 \ \rm{or} \
z=\pm\frac{1}{\sqrt{2}}\\
\end{eqnarray*}
Since $f(x,y,z)$ is positive at some points, negative at others, and
approaches $0$ as $(x,y,z)$ recedes to infinity, $f$ must have
maximum and minimum values at critical points.

Since $f(x,y,z)=0$ if $x=0$ or $y=0$ or $z=0$, the max and min values
must occur among the eight critical points where
\begin{eqnarray*}
x & = & \pm 1/\sqrt{2}\\
y & = & \pm 1/\sqrt{2}\\
z & = & \pm 1/\sqrt{2}
\end{eqnarray*}
At four of these points, $f$ has the value
$\frac{1}{2\sqrt{2}}e^{-3/2}$, the maximum value.

At the other four $f$ has the value $-\frac{1}{2\sqrt{2}}e^{-3/2}$,
the minimum value.

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