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\author{Kyle Spaans}
\date{June 19, 2009}
\title{Calculus 3 Lecture Notes}
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\section*{Lecture 20 -- Global Extrema and Optimizing With Contraints}
We finish the example from Wednesday, emphasising the importance of memorizing
trigonometric angles and values, which will be very important when we get to
triple integrals.
\paragraph*{Example 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
get values for $t$ of $\frac{\pi}{6}, \frac{5\pi}{6}, ...$ This give us a
value to plug into $f$ to findd the max: $f(\frac{\sqrt{3}}{2}, \frac{-1}{2})$
\paragraph*{Example 2}
Find the maximum and minimum values of
\[ f(x,y) 4xy - x^2 - y^2 - 6x \]
on the domain $S$ where $0 \le x \le 2, 0 \le y \le 3x$. Drawing this domain,
we get a triangle in quadrant 1, right angle at (2,0) and with a sloped side
made from the line $y = 3x$. $S$ is closed and bounded, $f(x,y)$ is continuous
by inspection, therefore the max and min exist. Now we can plug in our
algorithm.
\[ \nabla f = (f_x, f_y) = (4y - 2x - 6, 4x - 2y) = \vec{0} \]
And we can easily substitute $y = 2x$ into $f_x$ to find $x=1, y=2$ as a
critical point. Next, we examine each of the boundaries, considering them as
lines. Sub in $(x,0), (2,y), y=3x$ into $f$. This will give us two more
critical points in $S$: $(2,4), (\frac{3}{2}, \frac{9}{2})$. And finally check
the endpoints $(2,4), (2,0), (0,0)$. This lets us find the max 0 at $(0,0)$ and
min -16 at $(2,0)$.
\subsection*{Optimization with Contraints}
Find the maximum and/or minimum of $f(\vec{x})$ subject to constraint
$g(\vec{x}) = k$ (with $k$ constant). The constraint function gives us a curve
of points where the inputs $(x,y)$ give us output $k$. We assume there exists
a parameterization of $g(\vec{x})$: $x(t), y(t)$. Which gives us a nice
equation to try and optimize $f(x(t), y(t))$, finding critical points is as
simple as
\[ \frac{d}{dt} f(x(t), y(t)) = \vec{0} \]
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