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{\bf Exam Question

Topic: CriticalPoints}

Find and classify the critical points of the function
$$f(x,y)=y(x-3)^2-(x-y)(x+y-6).$$  \vspace{0.5in}

{\bf Solution}

$f(x,y)=y(x-3)^2-(x-y)(x+y-6),$ so equating the partial
derivatives to zero gives
\begin{eqnarray}
f_x&amp;=&amp;2y(x-3)-2x+6=0\\ f_y&amp;=&amp;(x-3)^2+2y-6=0
\end{eqnarray}

Equation (1) gives $2(x-3)(y-1)=0$ so $x=3$ or $y=1$. From
equation (2), $x=3$ gives $2y-6=0;\ \ y=3.$

Also from equation (2), $y=1$ gives $(x-3)^2=4,$ so $x=1$ or
$x=5.$

So the critical points are $(1,1),\ (3,3),\ (5,1).$

the second partial derivatives are $f_{xx}=2y-2,\ f_{yy}=2,\
f_{xy}=2(x-3).$ \vspace{0.3in}


\begin{tabular}{|c|c|c|c|c|c|}
\hline
  &amp;$f_{xx}$&amp;$f_{yy}$&amp;$f_{xy}$&amp;$\Delta=f_{xy}^2-f_{xx}f_{yy}$&amp;Type  \\
  \hline
 $(1,1)$&amp;0&amp;2&amp;$-4$&amp;$16&gt;0$&amp;SADDLE\\
 \hline
 $(3,3)$&amp;4&amp;2&amp;0&amp;$-8&lt;0$&amp;MIN $(f_{xx}&gt;0)$\\
 \hline
 $(5,1)$&amp;0&amp;2&amp;4&amp;$16&gt;0$&amp;SADDLE\\
 \hline
\end{tabular}
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