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 the 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​

Area

My initial attempt was to create an x and y table so that I could find different points that I could plot on the graph in order to see if it was going to be a line or parabola. In order to create the x and y table, I first plugged in a number for x and solved the equation. Let's use 0 for an example. When you plug 0 into the equation, it looks like this, y = 16-0². 0 squared is still 0. 16 - 0= 16. So the X value is 0 and the Y value is 16. So I kept plugging in different numbers for x in order to get a y value. Now that I have the x and y values, I'm able to multiply them together to find the area of the rectangles inside the parabola.

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Perimeter

My initial attempt was to plug in the x and y values into this equation to find the perimeter: 2(x+y).

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Solution

Largest Area

I knew that the biggest area of the rectangle was going to be between 2 & 3, so I tried numbers that were in-between those two numbers. I started off with 2.1 and worked my way up. As you can see on the left, the x value that got us the largest area was 2.3. In order to be more precise, I went to the nearest hundredth. So, I did the same process, I started off with 2.31 and worked my way up. After I found the largest area to the nearest hundredth, I stopped because my math teacher said that it was going to more time consuming and irrelevant if we go to the nearest thousandth. The value of x that got the largest area was 2.31​.

Largest Perimeter

I knew that the largest perimeter was going to be in-between 0 and 1, so I used 0.5 to see where that got me and moved my way up, and down if needed. As you can see on the left, I listed out x values that were between 0-1, and I also wrote what the perimeter would be. Therefore, the value of x that got the largest perimeter was 0.5​.

Group Test/Individual Test

 My group prepared for the group test by changing 16 into a different number, which would give us a new equation. The new equation was (y = 25 - x²). We used the same solving method as we did for the original method that would help us find the largest area and perimeter. I think my group did a good job on understanding the problem. We were always communicating with one another, and if one person was confused, someone would help them understand while the rest were trying to figure out the next step that we needed to complete in order to finish the problem. For the individual test, I did a good job but I made a tiny mistake but not reading the directions very carefully. The questions asked for the answer to the nearest tenth, and since I misread the question, I went to the nearest thousandth. Other than that small mistake, I think my understanding of the individual test was solid.

Evaluation/Reflection

What pushed my thinking was the perimeter because it was more of a process finding the biggest perimeter compared to area. What also pushed my thinking was when we came back from Thanksgiving break and we had to build off of other group's ideas to help us refresh our memory of the problem. This helped me because when we came back, I was lost on how to solve for perimeter, so I had to really think and look back at my notes to help myself remember. If I were to grade myself on this unit, I would give myself an A. The reason for this is because I think I had a good understanding of the problem  and I completed the work and learned the steps on how to solve for both biggest area and perimeter.