P.O.W. 2: The Gumball Dilemma
Graph 1
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Graph 2
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Graph 3
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Problem Statement
In this problem, one parent has a certain amount of children that all pass a gumball machine. This gumball machine has a certain variety of color. Each gumball costs a penny. We needed to find a formula that told us the maximum amount the parent would have to pay if all of his/her children wanted the same amount of gumballs. For example, Mrs. Hernandez has two twins who want the same color gumball.
Process Description/Solution
I started out small by calculating the maximum the parent would have to spend if (s)he had two kids. In Ms. Hernandez case, the most she'd have to pay is 3 cents. If there were three colors, the most she'd have to spend is 4 cents. I kept on calculating this until I had eight results. I then made a graph, as seen to the above, to the left.
I noticed that the colors and the cents correlated in their growth. To test if this would happen again, I tried doing the same process, but with triplets. As before, I used a table to organize my data. The first thing I noticed was that the correlation didn't exist when I used two children. However, I did find one formula to connect the information: C(K - 1) + 1 = ¢ (K = Kids, C= Children, and ¢ = Cents). This formula worked for both of the tables, so I decided to see if it'd continue to work if I used four children. It did work, which made me very confident in my theory, but I wanted to test one more time. Here is proof that the formula works for each table.
6(2 - 1) + 1 = 7
6(3 - 1) + 1 = 13
6(4 - 1) + 1 = 19
I decided to test it by using seeing what would happen if there 6 different colors and five children. The maximum amount a parent would pay would equal 25 cents. The equation went like this: 6(5 - 1) + 1 = 6(4) + 1 = 25. All of this evidence finally cemented my confidence in my formula.
I noticed that the colors and the cents correlated in their growth. To test if this would happen again, I tried doing the same process, but with triplets. As before, I used a table to organize my data. The first thing I noticed was that the correlation didn't exist when I used two children. However, I did find one formula to connect the information: C(K - 1) + 1 = ¢ (K = Kids, C= Children, and ¢ = Cents). This formula worked for both of the tables, so I decided to see if it'd continue to work if I used four children. It did work, which made me very confident in my theory, but I wanted to test one more time. Here is proof that the formula works for each table.
6(2 - 1) + 1 = 7
6(3 - 1) + 1 = 13
6(4 - 1) + 1 = 19
I decided to test it by using seeing what would happen if there 6 different colors and five children. The maximum amount a parent would pay would equal 25 cents. The equation went like this: 6(5 - 1) + 1 = 6(4) + 1 = 25. All of this evidence finally cemented my confidence in my formula.
Reflection
I learned that by being persistent, I will be able to find evidence that helps me prove or disprove my theories. I also learned that if different variables work inside the same constraints, there has to be a formula to connect them. I feel that I deserve a 10/10 on this project because I deeply explored the problem. I also managed to find the formula that showed how to find the maximum amount a parent would have to pay for the gumballs. The main reason I feel I deserve this grade is because I worked hard on this project and I really put a lot of time and effort into it.
The Habit of a Mathematician I used the most during this project was experiment through conjectures. I feel this because I invented a formula and tested if it worked with all of my data. Fortunately, I managed to hit the nail on the head early, so I didn't have to edit the formula as I went on, but, if I had to, I would've. I also tested it with a specialized case to see if my formula was valid.
The Habit of a Mathematician I used the most during this project was experiment through conjectures. I feel this because I invented a formula and tested if it worked with all of my data. Fortunately, I managed to hit the nail on the head early, so I didn't have to edit the formula as I went on, but, if I had to, I would've. I also tested it with a specialized case to see if my formula was valid.