\documentclass[12pt]{article}
\textwidth 6.5in\oddsidemargin 0in
\textheight 9in\topmargin -0.5in
\usepackage{amssymb}
\usepackage{amsmath}
\newtheorem{theorem}{Theorem}[section] 
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary} 
 \newtheorem{conjecture}[theorem]{Conjecture} 
\newtheorem{claim}[theorem]{Claim}
 \newtheorem{prop}[theorem]{Proposition}
\newtheorem{no}[theorem]{Notation} 
\newtheorem{question}[theorem]{Question}
\newtheorem{problem}{Problem} 
\newtheorem{definition}[theorem]{Definition}
\newtheorem{re}[theorem]{Remark} 
\newtheorem{pargrph}[theorem]{}
\newtheorem{examp}{Example}[section] 
\newtheorem{assume}{Assumptions}[section]
\newtheorem {exercise}[theorem] {Exercise} 


\newenvironment{remark}{\begin{re}\em}{\end{re}}
\newenvironment{example}{\begin{examp}\em}{\end{examp}} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\begin{center}
\large \bf Math 108\\
Final Solutions
\end{center}




\vspace{2mm}
\noindent
{\bf 1. } 
\begin{itemize}
\item[(a.)]  We solve $D(q) = S(q) $ or $\frac{125}{q + 1} = (q +
1)^2 $ or $(q+1)^3 = 125 $ or $q + 1 = 5$ or \fbox{$q_e = 4$}. 
Then $p_e = (4+1)^2 $ or  \fbox{$p_e = 25$}.
\item[(b.)]  The consumer surplus is
\[
CS = \int^4_0 \frac{125}{q + 1} dq - 4 \cdot 25 = 125 \ell
n(q+1)\big |^4_0 - 100 = \fbox{$125 \ell n 5-100$}.
\]
\item[(c.)]  The producer surplus is
\[PS = 100 - \int^4_0 (q+1)^2 dq = 100 - \frac{1}{3} (q + 1)^3
\big |^4_0 = 100 - \frac{125-1}{3} =
\fbox{$\displaystyle \frac{176}{3}$}.
\]
\end{itemize}


\vspace{2mm}
\noindent
{\bf 2. }   We have
\[\frac{\partial P}{\partial x} = 10 - 0.02 (2x + y) \mbox{ and }
\frac{\partial P}{\partial y} = 16 - 0.02 (x + 6y).
\]
To find the critical points, we solve $\frac{\partial P}{\partial
x} = 0 $ and $\frac{\partial P}{\partial y} = 0$.  This is $10 -
0.02 (2x + y) = 0 $ and $16 - 0.02 (x + 6y) = 0$.  Computing
gives $2x + y = 500 $ and $x + 6y = 800$.  Solving gives
\fbox{$x = 200 \mbox{ and } 6 = 100$}.  To use the second
derivative test we need
\[
\frac{\partial^2 P}{\partial x^2} = -0.04 \frac{\partial^2
P}{\partial y^2} = -0.12, \ \frac{\partial^2 P}{\partial x
\partial y} = -0.02.
\]
Then
\[
D = \left ( \frac{\partial^2 P}{\partial x^2} \right ) \left (
\frac{\partial^2 P}{\partial y^2} \right ) - \left (
\frac{\partial P}{\partial x \partial y} \right )^2 =
(-0.04)(-0.12) - (0.02)^2.
\]
Since $D = 0.0044 > 0 $ and $\frac{\partial^2 P}{\partial x^2}
= - 0.04 < 0,$ the values $x = 200 $ and $y = 10 $ maximize
the profit.  Then we compute that the maximum profit is
$P(200, 100) = 1,000.$ 




\vspace{5mm}
\noindent
{\bf  3.} ( {\bf Answer:} $q_e = 5, p_e = 25$ )





\end{document}
