**Problem Statement**

A rectangle has four points on a plane. The first point is on the origin, the second point is on the positive x axis, the third point is on the positive y axis, and the fourth point is on the graph of the equation y=16-x^2. The goal is to find the values of x (to the nearest hundredth) that yield the greatest area and the greatest perimeter.

**Process**

The sketch below is my initial representation of the problem. The constraints of the rectangle state that one corner must be at the origin, one at the positive x-axis, one at the positive y axis, and one on the graph y=16-x^2.

My first goal was to find the value of x that yields the greatest area. The equation for area, in this instance, is A=xy, or A=x(16-x^2). To find the x value that yields the greatest area, I simply began plugging x-values into this equation.

Here is a list of the x-values that I tried plugging-in:

x--A

0--0

1--15

2--24

3--21

4--0

x--A

0--0

1--15

2--24

3--21

4--0

By looking at the information above, I concluded that the x-value yielding the highest area must be between 1 and 3, since the area both reaches its maximum and begins decreasing all in-between these two x-values.

From there, I narrowed my search:

x--A

1--15

1.1--16.269

1.2--17.472

1.3--18.603

1.4--19.656

1.5--20.625

1.6--21.504

1.7--22.287

1.8--22.968

1.9--23.541

2--24

2.1--24.339

2.2--24.552

2.3--24.633

2.4--24.576

2.5--24.375

2.6--24.024

2.7--23.517

2.8--22.848

2.9--22.011

3--21

From this new set of data, I concluded that the x-value yielding the highest area must between 2.2 and 2.4. From there, I narrowed my search even further, since our assignment was to find the x-value to the nearest hundredth.

x--A

2.2--24.552

2.21--24.566

2.22--24.578

2.23--24.590

2.24--24.600

2.25--24.609

2.26--24.616

2.27--24.622

2.28--24.627

2.2924.631

2.3--24.6330

2.31--24.6336

2.32--24.6328

2.33--24.630

2.34--24.627

2.35--24.622

2.36--24.615

2.37--24.607

2.38--24.598

2.39--24.588

2.4--24.576

2.2--24.552

2.21--24.566

2.22--24.578

2.23--24.590

2.24--24.600

2.25--24.609

2.26--24.616

2.27--24.622

2.28--24.627

2.2924.631

2.3--24.6330

2.31--24.6336

2.32--24.6328

2.33--24.630

2.34--24.627

2.35--24.622

2.36--24.615

2.37--24.607

2.38--24.598

2.39--24.588

2.4--24.576

Based on this data, I found that the x-value yielding the highest area is 2.31, since the areas of the x-values adjacent to it are both lower than 24.6328.

The highest yielding x-value for area is 2.31.

After finding the x-value of the highest area, I had to find the x-value of the highest perimeter. The equation that I used to find the perimeter was P=2x + 2y, or P=2(x) + 2(16-x^2).

I then created a similar chart, but with x on the left and perimeter on the right.

x--P

0--0

1--32

2--28

3--20

4--0

x--P

0--0

1--32

2--28

3--20

4--0

By completing this part of the chart, I discovered that the highest-yielding x-value must be between 0 and 1, since the perimeters after that are continually decreasing. Now, I narrowed it down the x-values to the tenths place:

x--P

0--0

0.1--32.18

0.2--32.32

0.3--32.42

0.4--32.48

0.5--32.5

0.6--32.48

0.7--32.42

0.8--32.32

0.9--32.18

1--32

x--P

0--0

0.1--32.18

0.2--32.32

0.3--32.42

0.4--32.48

0.5--32.5

0.6--32.48

0.7--32.42

0.8--32.32

0.9--32.18

1--32

After making this chart, I realized that I actually didn't have to go any further. If you look, the perimeter value at x=0.5 divides the rest of the values symmetrically. This means that 0.5 is the highest possible x-value for the perimeter.

The highest-yielding x-value for the perimeter is 0.5.

**Group Test/Individual Test**

My group prepared for the group test by practicing some example problems in which we had to find the maximum yielding x-values for area and perimeter. Our goal was to practice the skills that we anticipated using on the test. For the individual test, I practiced plugging-in x-values because I sometimes make arithmetic errors when I have to do it so repetitively. In the end, I felt well-prepared for both the group test and the individual test because I felt knowledgeable on the subject and especially since we had a whole hour to answer only five-or-so questions.

**Reflection**

I didn't like this problem very much because a lot of the calculations required to solve it were extremely repetitive and boring. Although this guess-and-check method was very tedious, it was the only was to finish the problem. I think this problem gave us a lot of practice in plugging-in numbers into an equation, as well as relating the equation back to a geometric point-of-view. Combining parabolas with area/perimeter was a new experience for me. Overall, I think I should receive an A because I helped my group and collaborated with them on the problem.