\begin{frame} \frametitle{Optimization} \vspace{-1ex} \begin{exampleblock}{} A store sells 100 blu-ray players per week for $200$\$ each. A market survey shows that for each $10$\$ discount, the store would sell $40$ more players per week. The store buys the players at a price of $150$\$ per piece. \pause\smallskip What selling price would maximize the profit of the store? \pause\smallskip \begin{overlayarea}{\textwidth}{5cm} \only<-22>{ Introducing notation: \begin{itemize} \pause \item let $x$ be the discount \pause \item let $s$ be the number of players sold, and $p$ the profit \end{itemize}\vspace{-1ex} \pause \begin{talign} s(x) &= \mpause[1]{100 + 40\cdot \frac{x}{10}} \mpause{ = 100 + 4x}\\[-.5ex] \mpause{p(x) &= }\mpause{s(x) \cdot (200 - x - 150)} \mpause{ = (100 + 4x) \cdot (50-x)}\\[-.5ex] \mpause{&= -4x^2 + 100x + 5000} \hspace{1cm} \mpause{\text{for $x$ in $[\mpause[8]{0},\mpause[8]{50}]$}}\\[-.5ex] \mpause[9]{p'(x) &= -8x + 100} \hspace{1cm} \mpause{p'(x) = 0 \;\iff\; }\mpause{x=12.5} \end{talign} \pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause\pause Note that $p(x)$ is continuous\pause, and\vspace{-1ex} \begin{talign} p(0) &= \mpause[1]{5000} & p(12.5) &= \mpause[2]{5625} & p(50) = \mpause[1]{0} \end{talign} \pause\pause\pause } \only<23>{ \begin{center} \begin{tikzpicture}[default,scale=.65] \def\diaborderx{1.2cm} \def\diay{$p(x)$} \diagram{-0.5}{5}{-.5}{6}{1} \diagramannotatexx{1/10,2/20,3/30,4/40} \diagramannotateyy{1/1000,2/2000,3/3000,4/4000,5/5000} \diagramannotatez \begin{scope}[ultra thick] \draw[cgreen] plot[smooth,domain=0:5,samples=100] function{(100 + 4*10*x) * (50-10*x)/1000}; \end{scope} \end{tikzpicture} \end{center} } \end{overlayarea} By the Closed Interval Method, \alert{$12.5$\$ discount for maximal profit}.\hspace*{-10ex} \end{exampleblock} \end{frame}