\documentclass[12pt]{article} \pagestyle{empty} \setlength{\topmargin}{-1in} \setlength{\oddsidemargin}{-.3in} \setlength{\textheight}{9in} \setlength{\textwidth}{7.2in} \begin{document} \section*{Key for test 4} \begin{enumerate} \item $x$ is the speed and $t$ is the time. We want to know what is the speed when the time is 11 minutes. So, we want to know: what is $x$ when $t=11$. So we plug $t=11$ in and see what $x$ had to be. \begin{eqnarray*} 11 & = & .01(-597.62+.1976x^2+21.017x) \\ 0 & = & -5.9762+.001976x^2+.21017x - 11 \\ 0 & = & .001976x^2+.21017x -16.9762 \end{eqnarray*} With a zero on one side of the equation, we can now use the quadratic formula with $a=.001976$, $b=.21017$, and $c=-16.9762$. We store these numbers in our calculators or are very careful when entering them so that we don't get too much round off error. \begin{eqnarray*} x & = & \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ & = & \frac{-.21017 \pm \sqrt{(.21017)^2-4(.001976)(-16.9762)}}{2(.001976)} \\ & = & \frac{-.21017 \pm \sqrt{.1783513137}}{.003952} \\ & = & \frac{-.21017 \pm .422316603628}{.003952} \\ & = & \frac{.212146603628}{.003952} \mbox{ or } \frac{-.632486603628}{.003952} \\ & = & 53.6808207561 \mbox{ or } -160.042156788 \end{eqnarray*} If you are careful, you should get both of these answers exactly, with the possible exception of the last digit or 2. If you don't, you need to be more careful when entering on your calculator. Now of course, what does this mean? This is the speed after 11 minutes, and a speed of -160 doesn't make sense. So the answer is that after 11 minutes, the camper van is traveling at approximately 53.7 miles per hour. \item This was supposed to be fairly easy. We first look at the function and say,``This is a parabola, and it must be pointed down because the number in front of the $x^2$ is negative.'' What does that mean? It means that we have a vertex, and that vertex will have the biggest $y$ value of any point on the graph. We don't have a $y$, we have a $u$ and a $t$. So we look and see that the $u$ is playing the part of the $y$. Thus, we have a biggest $u$ value, or a maximum percentage of workers belonging to unions, and this is what we wanted to find. What year do we have our maximum? We use the formula that the $x$-coordinate of the vertex is at $x=\frac{-b}{2a}$. Again, we don't have $x$ and $y$, so we see that $t$ plays the role of $x$. So, our vertex is at $t=\frac{-b}{2a}=\frac{-2.38432}{2(-.022129)}=53.8731980659$. This is \emph{when} we have our maximum. What is the maximum percentage? We take this $t$ and use it to find $u$. So \begin{eqnarray*} u & = & -39.3213 + 2.38432(53.8731980659) - .022129 (53.8731980659)^2 \\ & = & 24.9041718062 \end{eqnarray*} So we have a maximum of 24.9\% when t=53.8 or close to 1954. Make sure that you understand how we found each of these parts and why the 53.87 is the year and the 24.9 is the percentage. You should also graph this function by plotting a bunch of points. \item First multiply it out to get $y=-2x^2-10x-13$. Notice that there is a $-2$ in front of the $x^2$, so the parabola is pointed down. Now where is the vertex? The $x$-coordinate of the vertex is at $x=\frac{-b}{2a}=\frac{10}{2(-2)}=-\frac{5}{2}$. Given an $x$ value, to find the $y$ value, we plug the $x$ into the original equation. So the $y$-coordinate of the vertex is $y=-2(-\frac{5}{2})^2-10(-\frac{5}{2})-13=-\frac{1}{2}$. So the vertex is at $(-\frac{5}{2},-\frac{1}{2})$ Now to graph it, we have most of the essential information. We know the coordinates of the vertex and that it is pointed down. This is already enough to get a pretty good idea of what the graph looks like. To get a better picture, we should plot some more points. \[ \begin{array}{r||r|r|r|r|r|r|r} x & -5 & -4 & -3 & -2 & -1 & 0 & 1 \\ \hline y & -13 & -5 & -1 & -1 & -5 & -13 & -25 \end{array} \] \item The first thing you need to do is figure out what you need to find. We need to find how many simple sticker and how many fancy stickers to make. Let $x$ be the number of simple stickers. Let $y$ be the number of fancy stickers. You could reverse these and everything will work out fine. Now in terms of hours, it takes $\frac{1}{4}$ hour to make a simple sticker and $\frac{1}{2}$ hour to make a fancy sticker. So if we make $x$ simple sticker and $y$ fancy ones, it takes us $\frac{1}{4}x+\frac{1}{2}y$ hours. We only have 30 hours, so we need to have \[ \frac{1}{4}x+\frac{1}{2}y \leq 30 \] We also can only make a total of 75 stickers so we need to have \[ x+y \leq 75 \] We also cannot make a negative amount of stickers so we also have $x \geq 0$ and $y \geq 0$. So we graph this system of inequalities, by first graphing the line. We need to be \emph{really} careful when graphing. We also find the corners. These should be: $(0,0),(75,0),(0,60),(30,45)$. We want to maximize the revenue. For us $R=2x+3y$ because we make \$2 for each simple sticker and \$3 for each fancy sticker. So we just plug the corners in and see which is the biggest. \[ \begin{array}{c||c|c|c|c} \mbox{corner} & (0,0) &(75,0) & (0,60) & (30,45) \\ \hline R & 0 & 150 & 180 & 195 \end{array} \] So we want to make $x=30$ and $y=45$, or 30 simple stickers and 45 fancy stickers. \end{enumerate} \newpage \section*{Homework} \begin{enumerate} \item Look at section 4.1 number 59 and answer: When is $R=80$? Find the vertex and explain the significance. \item Look at section 4.2 numbers 51-53 and answer: Find the vertex of the function and explain the significance. Find when $u(x)=0\%$. What values of $x$ make the model completely inaccurate and why. \item Change test problem 4. Now you raise the price of the stickers to \$3 and \$5. You become more efficient and can make the simple ones in 10 minutes and the fancy ones in 15 minutes, but you still only have 30 hours. You also expand the market so that you can now sell at most 150 stickers per week. How many should you make now? What if you charged \$4 for the simple and \$5 for the fancy? \end{enumerate} \end{document}