Problem Statement
A rectangle has one corner on the graph of y=16-x² , another at the origin, a third on the positive y-axis, and a fourth on the positive x-axis. If the area of the rectangle is a function of x, what value of x yields the largest area for the rectangle.
Process
Initial Attempts
Below this text to the left you can see my initial attempts for trying to find the maximum area and the maximum rectangle diagram. It has three possible rectangles which all landed on perfect points (whole numbers) the two that had the highest area were (2,12) and (3,7) which meant that our answer had to be around those points.
The next thing we decided to do to try and get a higher area was use Desmos (bottom middle) to get more exact points on the parabola you can see a random point in the picture and what we would do with a point like that in the graph was to plug it into the equation and see if the area got bigger or smaller. Finally to the right is our final diagram which includes the rectangle with the highest maximum area possible. To see how we got those points in the parabola scroll down to see the process.
The next thing we decided to do to try and get a higher area was use Desmos (bottom middle) to get more exact points on the parabola you can see a random point in the picture and what we would do with a point like that in the graph was to plug it into the equation and see if the area got bigger or smaller. Finally to the right is our final diagram which includes the rectangle with the highest maximum area possible. To see how we got those points in the parabola scroll down to see the process.
Solution
After we had both formulas we started to plug in numbers in attempt to find the maximum area and perimeter. The number we got for area was 2.31 and for perimeter was .5 we found them by plugging in numbers and finding the highest outcome for Y. Eventually we would end up finding a whole number where if you went up or down a number the outcome would be lower so we had to start using decimals. After playing around with decimals (tenths) we would find the highest possible outcome and that would be our answer.
Max Perimeter
P = -2x² + 2x + 32 - This is the formula we used to find the maximum perimeter. At first it may look confusing which is exactly how I felt but it is much more simple than it looks. The original formula for finding the perimeter of an object is P = 2L + 2w and in our case the w and L are the x and y (P = 2x + 2y). Since the y was given to us from the beginning all we had to do was plug all the components that are given to you and you end up with the max perimeter formula.
P = 2x + 2y / y=16-x² - P = 2x + 2 (16-x² ) = P = 2x² + 2x + 32
Side lengths: (.5 , 32.5)
P = 2x + 2y / y=16-x² - P = 2x + 2 (16-x² ) = P = 2x² + 2x + 32
Side lengths: (.5 , 32.5)
Max Area
A = 16x - x³ - This is the formula we used to find the maximum area and we came up with it exactly the same way we did with the max perimeter. We substituted the L and W with X and Y and plugged in the given which was the Y, so it looked something like this. A = L x W - A = X x Y - A = X (16-x² ) = A = 16x - x³
Final answer: (2.31 , 10.6639)
Final answer: (2.31 , 10.6639)
Group Test / Individual Test
The process of preparing for the group test went really well because we took as much advantage as we could when they gave us the time to practice with our group. We went through the problem as if it was the real test and everyone understood everything perfectly, so when it came to the test we were all prepared and based off the results the preparation paid off.
I was happy with my groups effort to understand the problem and I think we worked well together everyone was asking questions and we were very productive.
At first I struggled a bit with the two formulas which were for max area and perimeter since I wasn't here the days we discussed them but as soon as I got back I met with Mr. Carter to review how we came upon these formulas and as soon as he cleared that up for me I was back on track and understood everything perfectly. So for the test I understood everything on it and I was really happy with my results.
This group test was probably the best one I had in this semester because everyone seemed to understand it well and we were really productive. I had no negatives about this group test all positives.
I was happy with my groups effort to understand the problem and I think we worked well together everyone was asking questions and we were very productive.
At first I struggled a bit with the two formulas which were for max area and perimeter since I wasn't here the days we discussed them but as soon as I got back I met with Mr. Carter to review how we came upon these formulas and as soon as he cleared that up for me I was back on track and understood everything perfectly. So for the test I understood everything on it and I was really happy with my results.
This group test was probably the best one I had in this semester because everyone seemed to understand it well and we were really productive. I had no negatives about this group test all positives.
Reflection / Evaluation
During this problem what pushed my thinking the most was in the beginning when we had sketched out the parabola and plotted a few rectangles on the graph, I really didn't know where we were going to go from there. The part that I got the most out of happened to be the part that I struggled the most with which was understanding the two formulas and using them. Since I missed the day when we went over the formulas I was having a hard time understanding how we came up with them and how we used them but like I mentioned above as soon as I met with Mr. Carter I everything started to make sense. This taught me that it is very important to pay attention and that every part of a problem like this is very important. If I had to grade myself in this unit I would give myself an A because I felt like I worked hard and understood this unit very well. I also looked for help when I didn't understand something because I was absent which shows that I wanted to understand this unit well.