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Autor: anton 30 April 2011
Words: 1354 | Pages: 6
POW 17- Cutting the Pie
If you were given a pie what is the maximum number of pieces you can produce from 4, 5, and 10 cuts? Keep in mind, that the slices do not have to be the same size and the cuts do not necessarily have to go through the center of the pie, but the cuts do have to be straight and go all the way across the pie. Include any diagrams you used to find the solution such as an In-Out table, or any patterns you found.
The first thing I did to try to find my solution was to finish the In-Out table given, which already told us the maximum number of pieces that could be made with 1, 2 and 3 cuts. So I drew two circles, and drew in four cuts in one and five cuts in another to find the maximum number of pieces that could be produced. After several circles for each I found that the maximum number of pieces you can produce from four cuts is 11, and the maximum number of pieces for five cuts is 15.
In-X (Number of cuts) Out-Y (Maximum # of pieces)
So instead of trying to do the same thing t find out the maximum number of pieces for 10 cuts, I started looking for a pattern. I found that the difference between four and two is 2, the difference between seven and four is 3, the difference between eleven and seven is 4, and finally the difference between sixteen and eleven is 5. Based, on these results from my In-Out table I found out that one more value is added to the previous addend to come up with the next value. However, this wouldn't work that well because if I was to find the maximum number of pieces that can be produced from 50 cuts I would have to do a lot of tedious work to finally reach 50 cuts. This is because the independent variable is the number of cuts, not the maximum number of pieces. Finding the pattern was easy, but the challenging part was finding the formula. I tried finding something that had to do with the way the actual pie was cut, like how large the cuts had to be, or if it made a difference what direction the cut was in. Unfortunately, this did not help me in coming up with a solution or formula. So, I started looking at the difference between the number of cuts and the maximum number of pieces. I noticed that the range was increasing as the number of cuts increased. For a long time I plugged and chugged, playing around with multiplying, adding and squaring. I figured subtracting and dividing wouldn't really help since the numbers were increasing. Finally, I came to the conclusion that the number of cuts had to be squared and then another value had to be added. However, I realized this wouldn't work for only one cut because you couldn't add another value and have the maximum number of pieces be two. This led me to realize that at a certain point I had to divide. This had to be done after squaring the number of cuts and adding a value because dividing the number right after squaring it wouldn't work since not all the number of cuts and maximum number of pieces can be divided exactly by a smaller number. Still, there was something missing I needed the value in which I would add after squaring the number of cuts, the value in which I would later on divide. After awhile I came to the conclusion, that after squaring the number of cuts, I was to add the same number of cuts and divide by two. This gave me a very close answer but there was still something missing. I used 3 cuts to try to find what was missing. First, I squared 3 and got 9 then I added another 3 and received 12, when I divided that by two I received 6. One number short of my answer, so I decided that after dividing I was to add one to have the number 7 as my maximum number of pieces produced from three cuts. I did the same steps (xÐ’Ð† + x)/2] + 1) to four cuts and received 11 as the maximum number of pieces that can be produced. When, looking back at my In-Out table I realized that this was correct. I had finally found my formula:
[(xÐ’Ð† + x)/2] + 1
The largest number of pieces for 4 cuts is 11, the largest number of pieces for 5 cuts is 15, and finally the largest number of pieces for 10 cuts is 56. The formula I used to find these answers was (xÐ’Ð† + x)/2] + 1) . Steps done-
[(4Ð’Ð† + 4)/2] + 1= [(5Ð’Ð† + 5)/2] + 1= [(10Ð’Ð† + 10)/2] + 1=
(20/2) + 1= (30/2) + 1= (110/2) + 1=
10 +1=11 15 +1=16 55 +1=56
To make sure that you have the maximum number of pieces from any given number of cuts you could actually draw the cuts in the circles and count the amount of pieces or you could use the patter we found. Which was that one more value is added to the previous addend to come up with the next value. I think this pattern occurs because as the number of cuts increase so does the maximum number of pieces but even more than before. For example- the difference between four and two is 2, the difference between seven and four is 3, the difference between eleven and seven is 4, and finally the difference between sixteen and eleven is 5. If you continue to use this pattern you will also find that the maximum number of pieces from 4, 5, and 10 cuts is 11, 16, and 56 pieces.
In-X (Number of cuts) Out-Y (Maximum # of pieces)
Already we have begun to learn geometry ( the branch of mathematics that is concerned with the properties and relationships of points, lines, angles, curves, surfaces, and solids) using IMP. We already studying the properties and relationships of lines, and curves. I do not however see any investigational or explorative learning advantages from learning Geometry this way. I see a very large difficulty in learning this way because in a different situation I could have possibly never found a formula for my problem.
When grading my performance on a scale of 4 I believe that I deserve a four. When I first approached the problem I thought it was extremely difficult, I couldn't even notice a pattern! However, after sitting down and forcing myself to work on the problem, I started noticing a pattern. Once, that was set I started working on a formula, at first I thought this was impossible but after applying what I know and logic I slowly and reasonably came to find a formula. Although this took quite awhile, I believe that I did all this in a timely manner. By not waiting for the last minute to work on the problem but giving myself a lot of days I was able to ask help on anything I didn't understand from my peers. I also gave myself time to go over everything and do everything in an organized manner. I was not rushed and this allowed me to make any amends I felt was needed. I ended up solving the problem, finding a formula, actually understanding everything I did, all in a timely and organized manner. This is why, I believe that on a scale of 1 to 4 I believe that I deserve a four.
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