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

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

\textbf{Question}

\begin{description}
\item{(a)}
Find the critical points of the function $z=g(x,y)$ that satisfies the
equation $$e^{2zx-x^2}-3e^{2zy+y^2}=2.$$

\item{(b)}
Classify the critical points of $g$
\end{description}


\textbf{Answer}

\begin{description}
\item{(a)}
Differentiate the equation
$$e^{2zx-x^2}-3e^{2zy+y^2}=2.$$
with respect to $x$ and $y$, regarding $z$ as a function of $x$ and $y$:

\begin{eqnarray*}
e^{2zx-x^2}\left ( 2x \frac{\partial z}{\partial x} +2z-2x \right ) -
3e^{2zy+y^2}\left ( 2y \frac{\partial z}{\partial x} \right ) = 0 &amp; \
\ &amp; (*)\\
e^{2zx-x^2}\left ( 2x \frac{\partial z}{\partial y} \right )- 3e^{2zy+y^2}
\left ( 2y \frac{\partial z}{\partial y} +2z+2y \right ) =0 &amp; \ \ &amp;
(**)
\end{eqnarray*}
For a critical point we have $\frac{\partial z}{\partial x}=0$ and
$\frac{\partial z}{\partial y}=0$, and it follows from the equations
above that $z=x$ and $z=-y$.

Substituting these into the given equation we get:
\begin{eqnarray*}
e^{z^2}-3e^{-z^2} &amp; = &amp; 2\\
(e^{z^2})^2 - 2e^{z^2} - 3 &amp; = &amp; 0\\
(e^{z^2}-3)(e^{z^2}+1) &amp; = &amp; 0
\end{eqnarray*}
Thus $e^{z^2}=3$ or $e^{z^2}=-1$. 

Since $e^{z^2}=-1$ is not possible, we have $e^{z^2}=3$, so $z=\pm
\sqrt{\ln 3}$.

The critical points are ($\sqrt{\ln 3},-\sqrt{\ln 3}$) and
($-\sqrt{\ln 3}, \sqrt{\ln 3}$).

\item{(b)}
Use the second derivative test to classify the critical points found
in part (a). We need to calculate the second partials
\begin{eqnarray*}
A &amp; = &amp; \frac{\partial^2 z}{\partial x^2}\\
B &amp; = &amp; \frac{\partial^2 z}{\partial x \partial y}\\
C &amp; = &amp; \frac{\partial^2 z}{\partial y^2}
\end{eqnarray*}
to do this, differentiate (*) and (**) from (a).

Differentiating (*) with respect to $x$ gives
\begin{eqnarray*}
e^{2zx-x^2} \left [ \left ( 2x\frac{\partial z}{\partial x} +2z-2x
\right )^2 \right. &amp; &amp; \\
\left. +4\frac{\partial z}{\partial x}+ 2x \frac{\partial^2
z}{\partial x^2} -2 \right ] &amp; &amp; \\
-3e^{2zy+y^2}\left [ \left ( 2y \frac{\partial z}{\partial x} \right
)^2 + 2y\frac{\partial^2 z}{\partial x^2} \right ] &amp; = &amp; 0
\end{eqnarray*}
At a critical point, $\frac{\partial z}{\partial x}=0$, $z=x$, $z=-y$
and $z^2=\ln 3$, so
$$3 \left (2x \frac{\partial^2 z}{\partial x^2} -2 \right )
-\frac{3}{3} \left ( 2y \frac{\partial^2}{\partial x^2} \right ) =0$$
$$A=\frac{\partial^2z}{\partial x^2}=\frac{6}{6x-2y}$$

Differentiating (**) with respect to $y$ give
\begin{eqnarray*}
e^{2zx-x^2} \left [ \left ( 2x \frac{\partial z}{\partial y} \right
)^2 +2x \frac{\partial^2 z}{\partial y^2} \right ] &amp; &amp; \\
-3e^{2zy+y^2} \left [ \left ( 2y \frac{\partial z}{\partial y} +2z +2y
\right )^2 \right. &amp; &amp; \\
\left. + 4\frac{\partial z}{\partial y} +2y
\frac{\partial^2z}{\partial y^2} +2 \right ] &amp; = &amp; 0
\end{eqnarray*}
and evaluation at a critical point gives
$$3 \left ( 2x \frac{\partial^2z}{\partial y^2} \right ) - \frac{3}{3}
\left ( 2y \frac{\partial^2 z}{\partial y^2} + 2 \right ) = 0$$
$$C=\frac{\partial^2 z}{\partial y^2} = \frac{6}{6x-2y}$$

Finally, differentiating (*) with respect to $y$ gives
$$e^{2zx-x^2} \left [ \left ( 2x \frac{\partial z}{\partial x} +2z -2x
\right ) \left ( 2x \frac{\partial z}{\partial y} \right ) +2x 
\frac{\partial^2z}{\partial x \partial y} + 2
\frac{\partial z}{\partial y} \right ]$$
$$-3e^{2zy+y^2}\left [ \left (2y \frac{\partial z}{\partial y } +2z
+2y \right ) \left ( 2y \frac{\partial z}{\partial x} \right )
+2\frac{\partial z}{\partial x} +2y \frac{\partial^2z}{\partial
x}{\partial y} \right ]=0$$
and evaluating at a critical point,
$$(6x-2y)\frac{\partial^2}{\partial x}{\partial y} =0$$
so that
$$B=\frac{\partial^2}{\partial x \partial y} =0$$

At the critical point $(\sqrt{\ln 3}, -\sqrt{\ln 3})$ we have
\begin{eqnarray*}
A &amp; = &amp; \frac{6}{8\ln 3}\\
B &amp; = &amp; 0\\
C &amp; = &amp; \frac{2}{8\ln 3}
\end{eqnarray*}
so $B^2-AC&lt;0$, and $f$ has a local minimum at that critical point.

At $(-\sqrt{ln 3}, \sqrt{\ln 3})$ we have
\begin{eqnarray*}
A &amp; = &amp; -\frac{6}{8\ln 3}\\
B &amp; = &amp; 0\\
C &amp; = &amp; -\frac{2}{8\ln 3}
\end{eqnarray*}
so $B^2-AC&lt;0$ and $f$ has a local max at that point.

\end{description}

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