problem statement A rectangle has one corner on the graph of y=16-x^2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis If the area of the rectangle is a function of x, which value of x fields the largest area of the rectangle? Process and Solution When we were presented this problem the very first thing we had to do was write down questions we had about the question given. Things we didn't understand or wanted clarification on. After that we shared and answered them as a group.
My initial attempt was to try and factor the equation, for some reason. When we shared with our group our process a lot of us had empty pages and we weren't really sure what we were doing. After a while of brainstorming and factoring correctly, Mr. Carter helped us out by pointing at a graph and what we could do with it. Plot the points!
We started with the basic y and x intercepts, and made our parabola, then we sketched out a few rectangles inside the parabola. To find the largest area I initially found it the easiest way was just to count the squares inside the box. But it wouldn't give me the area. I knew which one it would be though. So we then plugged it into the equation. After plugging it in, we were a little stuck because we thought that was the final estimated area. We then realized that there could be super tiny points like 1.1, 1.2, 1.3, that could have the largest area instead of the whole number. So we created a x and y table. We stopped calculating them when the number began the become lower, so then we found the closest answer. The function we used to find this was, A=16x-x^3.
Our initial attempts to finding the perimeter was just plugging in the equating for perimeter: P=2(x+y). Then we simplified the function to, P=-2x^2+2x+32. We then found the max vertex which gave us the max perimeter. We new these functions were correct because it was giving us the correct answers when plotting them on the graph.
This problem is very basic math once you figure out the equations and how to plug in certain things into it. So as soon as we got the equation of what Y was equal too we figured out the Y intercept and the X intercepts after that we plugged in values for the different points for the parabola. After that we tried to draw the different rectangles within the parabola and ended up getting an x and y table that showed us the lengths and widths. After that we calculated the area for those which is just x times y. We saw that 2 had the largest area so decided to go along with that and try all of the numbers between 2 and 3. Instead of just doing x times y, we didn’t know the value of y we had to plug that into the equation so it was area=x(16-x^2) and we would plug in all of the different values of x in the equation. We distributed it out and thats how we got the equation for the area so it would be 16x-x^3. We saw that it started out small and then went up and then fell back down, that is because the equation is for a parabola. The point where it gets the biggest is at 2.3 as our x and 10.71 as our y. Since we found that out we went one step further using the same equation as the one before but plugged in all the numbers in between 2.3 and 2.4. We saw that at 2.31 is the largest that it got because at 2.32 it started to go down in the area. To find the perimeter of this problem we knew that the equation to find the perimeter is 2(x*y). We first plugged that into the regular x and y table and saw that 0 and 1 had the same perimeter so we would have to go in between them to be able to find the real perimeter. So we decided to go from .1 to .9. Plugging that into our new equation which is 2x+32-2x^2. The reason that it is different than the one before is because we had to find the value of Y and we plugged it into the value of Y and then distributed it out like how we did with the area. We find out that .5 gave us the largest perimeter and we even tried to do .51 but that was smaller than than 32.50 which was the largest perimeter. x=.5 y=15.75
Group Test/ Individual Test A day before our tests we prepared as a group by using the same problem but just changing the numbers in the function. y=__-x^2
At first when preparing we were very lost, we completely forgot about the whole process and how we found the solution. Little by little we started finding clues in our journals of how we found the maximum perimeter and area and it made sense by the end of the period. But, it took a while, don't get me wrong. It looked like we were all pretty much understanding it or remembering.
When we took the quiz as a group, it was a little unexpected. Everything was so confusing, we knew what was suppose to be done but there were little mistakes we came across that caused us to keep pausing. I think our group was just thinking about it way to much and we were very indecisive. By we, I mean me. I was so lost because I knew how to do it, but it was weird not being able to do it as a whole. The good thing was, we all communicated very well in what we were. doing to each other or what we didn't understand. We really weren't afraid to share our ideas.
It was so weird because when it came to the individual test, it was so easy. I probably did it all in 3 minutes. It was like engraved in my brain which was pretty cool because I knew what I was doing. I usually like group tests, I think it was just the over thinking that got me while the test. Overall I liked it, including both individual and group.