From 9d2cbd8679c846531a0325036a2e66524dd70420 Mon Sep 17 00:00:00 2001
From: Kyle Spaans
Date: Fri, 19 Jun 2009 10:58:22 -0400
Subject: [PATCH] almost done lec 20 notes
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+\documentclass{article}
+\usepackage{fullpage}
+\usepackage{amsmath}
+\author{Kyle Spaans}
+\date{June 19, 2009}
+\title{Calculus 3 Lecture Notes}
+\begin{document}
+\maketitle
+
+\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 $(x,0)$ ...
+
+\subsection*{Optimization with Contraints}
+Find the maximum and/or minimum of $f(\vec{x})$ subject to contraint
+$g(\vec{x}) = k$ (with $k$ constant)...
+\end{document}
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