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\title{MA162: Finite mathematics}
\subtitle{Introduction to Linear Algebra}
\author{Paul Koester}
\institute{University of Kentucky}
\date{March 12, 2014}

\begin{document}

\maketitle

\begin{frame}[t]{Linear Equations: Familiar Things}

In algebra, geometry, and calculus courses, linear equations are written in either
\begin{itemize}
  \item \textbf{Slope-Intercept Form:}  \( y = mx + b \)
  \vfill
  \item \textbf{Point-Slope Form:}  \(y = m(x - x_0) + y_0\)
  \vfill
\end{itemize}

Either of these forms are acceptable IF we want to think of \(y\) as being a function of \(x\),
as these forms tell us how \(y\) reacts to changes in \(x\).\\

If \(x_1 = x + \Delta x\) then 
\[ y_1 = m\cdot (x + \Delta x) + b = \left(mx + b\right) + m\cdot \Delta x = y + m\cdot \Delta x\]
In other words, if \(x\) increases by a fixed amount then \(y\) increases by \(m\) times that fixed amount.

\end{frame}


\begin{frame}[t]{Disadvantages of point-slope and slope-intercept forms}

\begin{itemize}
  \item Point-slope and slope-intercept forms are unable to handle vertical lines
  \vfill
  \item Point-slope and slope-intercept forms view one of the variables as a function of the other.
  In other words, the two variables are treated differently, whereas in business applications the variables 
  are generally on equal footing
  \vfill
  \item Generalizing point-slope and slope-intercept forms to handle more than two variables is awkward.
  \vfill
\end{itemize}
\end{frame}

\begin{frame}[t]{General Form of a Line}

The \alert{general form of a line} is an expression of the form
\[ Ax + By  = C \]
To be an equation of a line, we require that at least one of A or B is non-zero.\\
\vfill

\textbf{Example: }
Given the line \(6x + 2y = 4\), find 
\begin{itemize}
 \item \textbf{x-intercept}
 \vfill
 \item \textbf{y-intercept}
 \vfill
 \item \textbf{graph}
 \vfill
 \item \textbf{slope}
 \vfill
\end{itemize}

\end{frame}

\begin{frame}[t]{Special Cases of General Form}
\textbf{Vertical Lines} Suppose \(B = 0\) and \(A \neq 0\).  The line is then 
    \[  Ax + 0y = C \]
which is equivalent to \( x = \frac{C}{A}\), a vertical line.
\vfill
\textbf{Horizontal Lines} Suppose \(A = 0\) and \(B \neq 0\).
The line is then  \[  0x + By = C \]
which is equivalent to \( y = \frac{C}{B}\), a horizontal line.
\vfill
\textbf{Lines through Origin} Suppose \(A \neq 0\), \(B \neq 0\) and \(C = 0\).
    The line is then  \[  Ax + By = 0 \]
which is equivalent to \( y = - \frac{A}{B} x\), a line through the origin with slope \(-A/B\)
\vfill
\end{frame}

\begin{frame}[t]{Generic Case of General Form}
Consider a line with general form 
\[ Ax + By = C\]
in which none of A, B, and C are zero. 
Then
\begin{itemize}
 \item \textbf{x-intercept} is \((C/A, 0)\)
 \vfill
 \item \textbf{y-intercept} is \((0, C/B)\)
 \vfill
 \item \textbf{slope of y in terms of x}, \(\frac{\Delta y}{\Delta x} = -\frac{A}{B}\)
 \vfill
 \item \textbf{slope of x in terms of y}, \(\frac{\Delta x}{\Delta y} = -\frac{B}{A}\)
 \vfill
 \item To graph it, plot the x and y intercepts and connect the dots.
 \vfill
\end{itemize}

\end{frame}

\begin{frame}[t]{Identifying linear functions}
A stock analyst is looking at recent stock prices for several companies, and she wonders 
which of the prices involve a linear trend.
\vfill

\textbf{MNO stock:}

  \begin{tabular}{|l|l|l|l|l|l|}  \hline
  date	&  Feb 3 &    Feb 10   &  Feb 17  &  Feb 24 & Mar 3 \\ \hline
  price &  \$24.50 &  \$24.75  & \$25.00  & \$25.25 & \$25.50 \\ \hline
  \end{tabular}
\vfill

\textbf{PQR stock:}

  \begin{tabular}{|l|l|l|l|l|l|}  \hline
  date	&  Feb 3 &    Feb 10   &  Feb 17  &  Feb 24 & Mar 3 \\ \hline
  price &  \$32.00 &  \$31.00  & \$29.00  & \$28.00 & \$25.00 \\ \hline
  \end{tabular}
\vfill

\textbf{STU stock:}

  \begin{tabular}{|l|l|l|l|l|}  \hline
  date	&  Feb 3 &    Feb 10   &  Feb 24   & Mar 10  \\ \hline
  price &   \$80 &    \$82     &   \$86    &  \$90    \\ \hline
  \end{tabular}
\vfill
\end{frame}

\begin{frame}[t]{Interpreting the Coefficients}
You manufacture pencils.  
You observe that you will sell 4000 pencils per month if you charge \(\$0.50\) per pencil, 
and 3700 per month if you charge \(\$0.65\) per pencil.
Let \(q(s)\) denote the number of pencil you expect to sell at price \(s\).
\begin{itemize}
  \item Assuming \(q(s)\) is linear, find an expression for \(q(s)\).
  \vfill
  \item In plain english, explain the meaning of the \(s\)-intercept, \(y\)-intercept, and slope.
  \vfill
  \item You decide to make 3800 pencils.  What is the highest price you can charge if you want to sell all 3800 pencils?
  \vfill
\end{itemize}
\end{frame}

\begin{frame}[t]{Cost}  
Suppose the cost to produce \(q\) pencils per month is \(C(q) = \$0.30 q + \$1200 \).
\begin{itemize}
  \item Find the \(y\)-intercept and explain its meaning. 
  \vfill
  \item Find the slope and explain its meaning.
  \vfill
  \item Determine the cost to produce 3800 pencils.  
  \vfill
  \item Is this a good business plan?
  \vfill
\end{itemize}

\end{frame}



\end{document}
