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\begin{document}

\title{Econ 340: Week 9}



\author[1]{Ennio E. Piano}%
\affil[1]{George Mason University}%


\vspace{-1em}



  
  \date{May 08, 2024}


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\sloppy


\section*{A note on the effect of per-unit taxes on supply and
demand}

{\label{878937}}

\subsection*{Supply side}

{\label{572767}}

\begin{quote}
\(D:\ Q_d=a-bP \)

\(S:\ Q_s=\alpha+\beta(P-t)\)

Equilibrium

\(a-bP=\alpha+\beta(P-t)\)

\((\beta+b)P=a+\alpha+\beta t\)~

\(P^{*}=\frac{a+\alpha+\beta t}{\beta+b}\)

\(Q^{*}=a-b\frac{a+\alpha+\beta t}{\beta+b}\)

Taking the first derivative, we can show that an increase in taxation
increases the equilibrium price level, while the equilibrium quantity
falls
\end{quote}

\subsection*{Demand side}

{\label{414830}}

\begin{quote}
\(D: Q_d=a-b(P+t)\)

\(S:\ Q_s=\alpha+\beta P\)

Equilibrium

\(a-b(P+t)=\alpha+\beta P\)

\(P(\beta+b)=a-\alpha-bt\)

\(P^{*}=\frac{a-\alpha-bt}{\beta+b}\)

\(Q^{*}=\alpha+\beta\frac{a-\alpha-bt}{\beta+b}\)

Taking the first derivative, we can show that the equilibrium price is
decreasing in the tax, while the equilibrium quantity is decreasing in t
\end{quote}

\section*{Single variable
optimization}

{\label{828827}}

\subsection*{Intro}

{\label{735625}}

\begin{itemize}
\tightlist
\item
  Maxima and minima are both~\emph{extreme points}
\item
  Definition of a maximum:~\(c\in D\) is a maximum point
  for~\(f\),~\(\iff f(x)\leq f(c)\ \forall\ x\in D\)
\item
  Definition of a minimum:~\(d\in D\) is a minimum point
  for~\(f,\ \iff f(x)\geq g(d)\ \forall\ x\in D\)
\item
  Thus, f(c) is the~\emph{maximum value} of the function, while f(d) is
  its~\emph{minimum value}
\item
  Strict maxima and minima:~\(\leq \&amp; \geq\) now
  become~\(< \&amp; >\)
\end{itemize}

\subsubsection*{Example 1}

{\label{408928}}

\begin{quote}
Find the maximum of the function~\(f(x)=3-(x-2)^2\)

Remember that~\((x-2)^2\) is positive for any value of x. Thus,
the function must be maximizes at~\(f(x)=3\)

Does the function have a minimum?

What happens to the function as x goes to~\(\infty\)?
\end{quote}

\subsubsection*{Example 2}

{\label{947272}}

\begin{quote}
\(g(x)=\sqrt{x-5}-100\)

Remember that x must be larger than 5, which means that the lowest value
the function can take is~\(f(x)=100\)

Does the function have a maximum?
\end{quote}

\begin{itemize}
\tightlist
\item
  Stationary points: If f(x) is differentiable with both a minimum and a
  maximum at some interior point c of its domain, the tangent line must
  be horizontal. Unfortunately, this is a necessary but not a sufficient
  feature for a maximum or a minimum. This is because the same is true
  for~\emph{all~}stationary points.
\item
  Formally, if a function is differentiable over the
  interval~\(I\) and~\(c\) is an interior
  point of~\(I\) and~\(f^{'}(x)=0\)
  for~\(x=c\), then~\(c\) is a stationary
  point
\end{itemize}

\subsubsection*{Proof}

{\label{855782}}

\begin{quote}
Assume~\(c\) is a maximum point of~\(f\)
over the interval~\(I\). Since~\(c\) is an
interior point of~\(I\), then there exist
an~\(h\) such that~\(c+h\in I\)

By definition of a maximum,~\(f(c+h)\leq f(c)\), which
implies~\(f(c+h)-f(c)\leq 0\)

For a small~\(h\) enough, we can write Newton's quotient

\begin{quote}
\(\frac{f(c+h)-f(c)}{h}\leq 0\)
\end{quote}

The limit for~\(h\to 0\) from the right yields:

\begin{quote}
\(\lim_{h\to 0}^{+}\frac{f(c+h)-f(c)}{h}= 0\)~
\end{quote}

The limit for~\(h\to 0\)~from the left yields:

\begin{quote}
\(\lim_{h\to 0}^{-}\frac{f(c+h)-f(c)}{h}= 0\)
\end{quote}

Thus,~\(f^{'}(c)=0\)

QED
\end{quote}

\subsection*{Simple tests for extreme
points}

{\label{787249}}

\begin{itemize}
\tightlist
\item
  In some cases, we can find the maximum or minimum values of a function
  just by studying the sign of its first derivative
\item
  Suppose~\(f(x)\)~is differentiable in an
  interval~\(I\) and that is has only one stationary
  point,~\(x=c\). Suppose~\(f^{'}(x)\ge 0\ \forall x \in I\) such
  that~~\(x\leq c\), whereas~\(f^{'}(x)\leq 0\ \forall x\in I\) such
  that~\(x\ge c\). Then,~\(f(x)\) is increasing to
  the left of~\(c\) and decreasing to the right
  of~\(c\). It follows that~\(x=c\) is a
  maximum point for f in the interval
\end{itemize}

\subsubsection*{Example 1}

{\label{656291}}

\begin{quote}
\(c(t)=\frac{t}{t^{2}+4}\),~\(t\ge 0\)

Find the maximum of c(t)

\(\frac{dc(t)}{dt}=\frac{(t^{2}+4)-(t)(2t)}{(t^{2}+4)^{2}}=\frac{4-t^{2}}{(t^{2}+4)^{2}}=\frac{(2-t)(2+t)}{(t^{2}+4)^{2}}\)

This derivative would be positive for~\(t<2\) and negative
for~\(t>2\). Thus,~\(t=2\) is the maximum of the
function
\end{quote}

\subsubsection*{Extreme points for concave/convex
functions}

{\label{563498}}

\begin{itemize}
\tightlist
\item
  If~\(f(x)\) is concave, that is,~\(f^{''}(x)\leq 0\), over
  the interval~\(I\), and the function contains a
  stationary point~\(c\) in~\(I\),
  then~\(c\) is a maximum
\item
  If~\(f(x)\) is convex, that is,~\(f^{''}(x)\geq 0\), over
  the interval~\(I\), and the function contains a
  stationary point~\(c\) in~\(I\),
  then~\(c\) is a minimum
\end{itemize}

\subsubsection*{Example 3}

{\label{353088}}

\begin{quote}
\(f(x)=e^{x-1}-x\)

\(f^{'}(x)=e^{x-1}-1=0\)

\(e^{x-1}=1\)

\((x-1)\ln{e}=\ln{1}\)

\(x-1=0\Longrightarrow x=1\)

\(f^{''}(x)=e^{x-1}>0\)

Thus, the function is convex, which makes~\(x=1\)~the
minimum
\end{quote}

\subsection*{Economic examples}

{\label{414539}}

\subsubsection*{Example 1}

{\label{771020}}

\begin{quote}
Suppose~\(Y(N)\) bushels of wheat are harvested per acre of
land when N pounds of fertilizers per acre are used. If~\emph{P~} is the
dollar price per bushels of wheat and~\emph{q} is the dollar price per
pound of fertilizer, then profits in dollars per acre are

\(\pi(N)=PY(N)-qN,\ N\geq0\)

Suppose there exists N* such that~\(\pi^{'}(N)\geq 0\),
for~\(N\leq N^{*}\), whereas~\(\pi^{'}(N)\leq0\)
for~\(N\geq N^{*}\). Then,~\(N^{*}\) maximizes profits
and~\(\pi^{'}(N)=0\). That is,~\(PY^{'}(N^{*})=q\)

Suppose the function~\(Y(N)=\sqrt{N}\),~\(P=10\),
and~\(q=.5\).

\(10\cdot .5 N^{-.5}=.5\)

\(N^{-.5}=.1\)

\(N^{*}=100\)
\end{quote}

\subsubsection*{Example 2}

{\label{182613}}

\begin{quote}
\(\)a)~\(c(q)=2q^{2}+10q+32\)

b)~\(c(q)=aq^2+bq+c\)
\end{quote}

\subsubsection*{Example 3}

{\label{420099}}

\begin{quote}
A monopolist is faced with the demand function denoting the
price~\(P(Q)\)~when output is~\(Q\). The
monopolist has a constant average cost~\(k\) per unit
produced.

a) Find the profit function~\(\pi(Q)\), and prove that the FOC
for maximizing profits is

~~~~\(P(Q^{*})+Q^{*}P^{'}(Q^{*})=k\)

b) By implicit differentiation, find how the monopolist's choice of
optimal is affected by changes in~\(k\)

c) How does the optimal profit react to a change
in~\(k\)?
\end{quote}

\subsection*{The extreme value theorem}

{\label{452440}}

\begin{itemize}
\tightlist
\item
  Important question: How can we locate extreme points when the function
  is not steadily increasing and/or decreasing?
\item
  The extreme value theorem:
\end{itemize}

\begin{quote}
If~\emph{f} is continuous function over a closed, bounded interval
{[}\emph{a, b}{]}, then there exists a point~\emph{d} in {[}\emph{a,
b}{]} where~\emph{f} has a minimum, and a point~\emph{c} in {[}\emph{a,
b}{]} where~\emph{f} has a maximum, so that~\(f(d)\leq f(x)\leq f(c)\ \forall\ x \in [a, b]\)
\end{quote}

\begin{itemize}
\tightlist
\item
  The conditions of the extreme value theorem are~\emph{sufficient} but
  not~\emph{necessary}
\end{itemize}

\subsubsection*{How to find minima and
maxima}

{\label{871986}}

\textbf{Problem:~}Find the maxima and minima of a differentiable
function~\emph{f} defined on a closed, bounded interval {[}\emph{a,
b}{]}

\textbf{Solution:}

\begin{enumerate}
\tightlist
\item
  Find all stationary points of~\emph{f} in (\emph{a, b})-- that is,
  find all points~\emph{x} in (\emph{a, b}) that satisfy the
  equation~\emph{f'(x)=0}
\item
  Evaluate~\emph{f} at the end points~\emph{a} and~\emph{b} of the
  interval and also at all stationary points
\item
  The largest function value found in 2. is the maximum value and the
  smallest function value is the minimum value of the function
\end{enumerate}

\subsubsection*{Example 1}

{\label{888356}}

\begin{quote}
Find minimum and maximum of~\(f(x)=3x^{2}-6x+5\),~~\(x\in [0, 3]\)
\end{quote}

\subsubsection*{Example 2}

{\label{476910}}

\begin{quote}
Find the minimum and maximum value
of~\(f(x)=.25x^{4}-\frac{5}{6}x^{3}+.5x^{2}-1\),~~\(x\in [-1, 3]\)
\end{quote}

\subsubsection*{The mean-value theorem}

{\label{166879}}

Skip

\subsection*{Further economic examples}

{\label{422320}}

\subsubsection*{Example 1}

{\label{143215}}

\begin{quote}
\(\pi(Q)=R(Q)-C(Q)\)

FOC

\(R^{'}(Q^{*})=C^{'}(Q^{*})\)

Thus,~\emph{production should be adjusted to a point where the marginal
revenue is equal to the marginal cost}.
\end{quote}

\subsubsection*{Example 2}

{\label{797339}}

\begin{quote}
Do example 2, p. 288
\end{quote}

\subsubsection*{Example 3}

{\label{727959}}

\begin{quote}
Neither a borrower nor a lender be,

For loan oft loses both itself and friend,

And borrowing dulls the edge of husbandry.

\href{https://www.owleyes.org/text/hamlet/read/act-i-scene-iii/root-71604-12/75707}{Hamlet
Act 1, scene 3, 75--77}
\end{quote}

\begin{quote}
The intertemporal choice problem: how should you allocate consumption
over time (say, two periods)?

Assume a log utility function:

\(U(c_{1},c_{2})=\ln{c_1}+\frac{1}{1+\rho}\ln{c_{2}} \)

Where you can only spend on~\(c_2\) what you have not spent
on~\(c_1\), or

\(c_{2}=y_{2}-(1+r)(c_{1}-y_{1})\)

Substituting into the utility function

\(U(c_{1},c_{2})=\ln{c_1}+\frac{1}{1+\rho}\ln{[y_{2}-(1+r)(c_{1}-y_{1})]} \)

FOC

\(\frac{1}{c_1}+\frac{1}{1+\rho}\frac{1}{y_{2}-(1+r)(c_{1}-y_{1})}\cdot (-(1+r))=0\)

Solving for~\(c^{*}_{1}\)

\(c^{*}_{1}=\frac{y_{2}+ry_{1}}{2+\rho}\)
\end{quote}

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