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.
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.
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.
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.
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.