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\author{Kyle Spaans}
\date{June 17, 2009}
\title{Calculus 3 Lecture Notes}
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\section*{Lecture 19 -- Global Extrema}
When do global extrema exist? Given $f(x,y)$ on domain $D$, what conditions do
we need to impose on $f$, $D$ such that max and min of $f$ exist? First
consider simpler examples with $y = f(x)$. If $y = \frac{1}{x}$ and $x > 0$
there is no minimum because $D = (0,\infty)$ is unbounded on the right and no
maximum because the left side of the domain is open (imagine the graph of this
function). Another function that is discontinuous, it has no max.
\paragraph*{Theorem}
A continuous function $f$ on a closed and bounded domain $D$ reaches its max
and min. (Note that is sufficient, but not a necessary condition. We'll see an
example of a discontinuous function that has a minimum.)
Returning to functions of two variables now, things are very similar. $f(x,y) =
x^2 + y^2$ on $R^2$ % How to get the ``real'' symbol again?
we can see that there is no max, because $R^2$ is not bounded above. The min is
$(0,0)$. If we make the function discontinuous by defining $f(0,0) =
\frac{1}{2}$ then there would be no minimum anymore. However, if we define it
to be $-\frac{1}{2}$ instead, then the discontinuous function will have a min
of $-\frac{1}{2}$.
\emph{Definitons:}
\begin{itemize}
\item A point $\vec{a} \in D$ is a \textbf{boundary point} if any neighbourhood
of $\vec{a}$ contains points belonging to $D$ and points not belonging to
$D$.
\item A set of all boundary points of $D$ forms a \textbf{boundary} of $D$.
\item If $D$ contains its boundary, it is \textbf{closed} (otherwise it is
open).
\item $D$ is bounded if there is a ball of radius $r$ centered at $\vec{a}$
such that $D$ belongs to this ball ($\| \vec{x} - \vec{a} \| \le r$).
\end{itemize}
The theorem for maxima and minima is the same for functions of two variables.
\paragraph*{Algorithm for finding global extrema}
We assume $f$ is continuous and that $\nabla f$ exists.
\begin{enumerate}
\item Compute $\nabla f(x) = 0$ (find local extrema)
\item Check the boundary of $D$ (in function of two variables the boundary can
be a curve)
\begin{enumerate}
\item Parameterize the boundary of $D$ ($x(t), y(t)$).
\item Substitute in $f(x(t), y(t))$
\item Compute $\frac{d}{dt} f(x(t), y(t)) = 0$ (find all critical points
on the boundary)
\item If the boundary has corners, add them to the list of critical
points.
\end{enumerate}
\item Check all points for max and min values.
\end{enumerate}
\paragraph*{Example 1}
Let $f(x,y) = 100 + 10(3x^2y - y^3)$, $D: x^2 + y^2 \le 1$. Find the max and
min of $D$. To solve, use the algorithm. Find
\[ \nabla f(x) = (60xy, 10(3x^2 - 3y^2)) \]
and where is it equal to 0? We can see that is it when $x,y = 0$. Substitute
$x = 0$ into $10(3x^2 - 3y^2)$ to get that $y = 0$ again. Therefore the only
critical point is $(0,0) \in D$. Next, examine the boundary $x^2 + y^2 = 1$.
Parameterize $x = \cos t$, $y = \sin t$. Compute
\[ \frac{d}{dt} f(x(t), y(t)) = 30\cos t (\cos^2 t -3\sin^2 t) = 0\]
And solving that to find where it is equal to 0 will give us the extrema. We
will solve the rest of this on Friday.
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