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%\fancyhead[CE]{\fontsize{9}{11}\selectfont F. GOTTI AND M. GOTTI}
%\fancyhead[CO]{\fontsize{9}{11}\selectfont ATOMICITY AND BOUNDEDNESS OF MONOTONE PUISEUX MONOIDS}
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\begin{document}
	
%	\mbox{}
%	\title{Limits and Derivatives (MATH 53)}
%	
%	\maketitle



\section*{Maxima, Minima, and Lagrange Multipliers}
\bigskip

\begin{problem}
	Find a function $f \colon \rr^2 \to \rr$ whose set of critical points is: \\
	
	\hspace{20pt} (1) $\rr^2$ \quad \quad (2) $\{x,y \mid xy =0\}$ \quad \quad (3) $\zz \times \zz$
\end{problem}

\begin{problem} [Cal Final, Summer 2018W]
	Find the minimum and maximum values of the function $f(x,y,z) = x+y+z$ subject to the constraint $x^2 + y^2 + z^2 \le 10$.
\end{problem}

%\noindent \underline{Solution:}
%	(1) First let us compute the slope:
%	\[
%		\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3e^{3t}}{e^{-t}(1-t)} = \frac{3e^{4t}}{1-t}.
%	\]
%	At $x = 0$, we have that $t = 0$. Therefore $(dy/dx)(0) = 3$, which implies that the equation of the line is $y = 3x + 1$.
%	
%	(2) The curve is vertical when $dx/dt = 0$ and $dy/dt \neq 0$. Then $e^{-t}(1-t) = 0$, which implies $t=1$. As $(dy/dt)(1) \neq 0$, it follows that the only point where the curve is vertical is $(e^{-1}, e^3)$ when $t=1$.
%\hfill \qed

%\begin{problem}
%	Find the length of the curve
%	\[
%		x = e^t + e^{-t}, \quad \quad y = 2018 + 2t.
%	\]
%\end{problem}
%
%\noindent \underline{Solution:}
%	Since $dx/dt = e^t - e^{-t}$ and $dy/dt = 2$, the length of the curve is given by
%	\[
%		\int_0^1 \sqrt{(e^t - e^{-t})^2 + 4} \, dt = \int_0^1 \sqrt{e^{2t} + 2 + e^{-2t}} \, dt = \int_0^1 (e^t + e^{-t}) \, dt = e - e^{-1}.
%	\]
%\hfill \qed

%\section{Polar Coordinates}
%
%\begin{problem}
%	Suppose that the polar curve $r = f(\theta)$, where $0 \le \theta \le 2\pi$ has length $L$ and encloses a region of area $A$.
%	
%	\begin{enumerate}
%		\item What is the area of the region enclosed by the polar curve $r = 4 f(\theta)$, where $0 \le \theta \le 2\pi$?
%		\item What is the length of the polar curve $r = 5 f(\theta)$, where $0 \le \theta \le 2\pi$?
%	\end{enumerate}
%\end{problem}

%\noindent \underline{Solution:}
%	Let $A'$ be the area of the region enclosed by the polar curve $r = 4 f(\theta)$, where $0 \le \theta \le 2\pi$, and let $L'$ be the length of the polar curve $r = 5 f(\theta)$, where $0 \le \theta \le 2\pi$. Then
%	\[
%		(1) \quad \quad \quad \quad A' = \int_0^{2\pi} \frac 12 [4f(\theta)]^2 \, d\theta = 16A 
%	\]
%	and
%	\[
%		(2) \quad L' = \int_0^{2\pi} \sqrt{[5 f(\theta)]^2 + \big[\frac{d(5f(\theta))}{d\theta}\big]^2} \, d\theta=  5\int_0^{2\pi} \sqrt{[f(\theta)]^2 + \big[\frac{d(f(\theta))}{d\theta}\big]^2} \, d\theta = 5L.
%	\]
%\hfill \qed
%
%\begin{problem}
%	\begin{enumerate}
%		\item Sketch the region that lies inside both of the polar curves $r = 1 + \cos \theta$ and $r = 1 - \cos \theta$.
%		\item Compute the area of the region.
%	\end{enumerate}
%	
%\end{problem}

%\noindent \underline{Solution:}
%	\begin{enumerate}
%		\item The red part of the figure shows the desired enclosed region:
%		\begin{center}
%			\begin{figure}[h]
%				\includegraphics[width=10cm]{images/AreaBetweenCosines.JPG} \\
%				\caption{Colored red, the region between $r = 1 + \cos \theta$ and $r = 1 - \cos \theta$.}
%			\end{figure}
%		\end{center}
%		\vspace{5pt}
%		\item By symmetry, we that the area enclosed by the the two curves is
%		\[
%			4 \int_0^{\pi/2} \frac 12 \big(1 - \cos \theta \big)^2 \, d\theta = \frac{3 \pi}{2} - 4.
%		\]
%	\end{enumerate}
%	
%\hfill \qed
	
\vspace{20pt}

\begin{thebibliography}{20}
	
	\bibitem{jS15} J.~Stewart: \emph{Calculus} 8th Edition, Cengage Learning, Boston 2016.
	
\end{thebibliography}
	
\end{document}