A normal can of Coca Cola carries about 365.3 mL of soda and uses about 296.1 cm squared of aluminum to make the can. While this is an ideal amount of liquid, the cost of each can may be reduced by adding a few adjustments to the size of the can and the amount of aluminum used. The cost of one standard can of Coca Cola to be manufactured is $0.047, which adds up. Coca Cola produces about 19 billion cans per year and that costs about 893 million dollars. As a class, we looked at how we can reduce the cost of a Coca Cola can and which is the cheapest can by testing different heights and radiuses.
To start, we looked at a standard can of Coca Cola by finding the volume, which established a base line. We tried to create a can that was less money, but still held the same amount of liquid. For my first test, I decided to double the height of a regular can by changing the height to 24.2cm. By using the equation: V = π • r-squared • h (r=radius V=volume h=height), I was able to find the accurate radius of the can.
To start, we looked at a standard can of Coca Cola by finding the volume, which established a base line. We tried to create a can that was less money, but still held the same amount of liquid. For my first test, I decided to double the height of a regular can by changing the height to 24.2cm. By using the equation: V = π • r-squared • h (r=radius V=volume h=height), I was able to find the accurate radius of the can.
365.3 = π • r-squared • 24.2
365.3/2 = π • r-squared • 24.2/24.2
14.3/π = π • r-squared/π
√4.6 = √r-squared
2.1 = r
365.3/2 = π • r-squared • 24.2/24.2
14.3/π = π • r-squared/π
√4.6 = √r-squared
2.1 = r
I tried different measurements, this time doubling the radius to 6.2cm. Then, I used the same equation to find the height of the can.
365.3 = π • 6.2-squared• h
365.3/π = π • 6.2-squared • h/π
116.3 = 38.44 • h
116.3/38.44 = 38.44/38.44 • h
3.03 = h
365.3/π = π • 6.2-squared • h/π
116.3 = 38.44 • h
116.3/38.44 = 38.44/38.44 • h
3.03 = h
After all these calculations, I put the measurements into a table to compare all the data.
Once the table was completed, I had to find the Surface Area (S.A.) for each can and the cost to see how much more or less efficient the can if. The equation to find the surface area is: S.A. = (2 • π • r-squared) + (2 • π • r • h). The cost of the can is: Cost = S.A. • 0.00016. First, I calculated the surface area and cost for the standard Coca Cola can.
S.A. = (2 • π • 3.1-squared) + (2 • π • 3.1 • 12.1)
S.A. = 60.4 + 235.7 S.A. = 296.1cm-squared |
Cost = 296.1 • 0.00016
Cost = $0.047376 |
Next, I found the surface area and cost for can #1 and can #2.
Can #1
S.A. = (2 • π • 2.1-squared) + (2 • π • 2.1 • 24.2)
S.A. = 27.7 + 319.3 S.A. = 347.02 cm-squared |
Cost = 347.02 • 0.00016
Cost = $0.0555232 |
Can #2
S.A. = (2 • π • 6.2-squared) + (2 • π • 6.2 • 3.03)
S.A. = 241.5 + 118.04 S.A. = 359.6 cm-squared |
Cost = 359.6 • 0.00016
Cost = $0.057536 |
Here is a table to compare the data.
We compiled all the data we acquired as a class and put it into one graph. The data looks like a stretched out "U" and, at the very bottom, is the least expensive can for Coca Cola with the lowest surface area and cost. To double check this, we created a formula to find surface area in terms of radius. The equation is: S.A. = 2 • π • r-squared + 2 (V/r). The radius for the least expensive can would be about 3.8 cm. By using the equation to find surface area, I was able to find that the surface area of the can is 282.7 cm-squared. The cost of this can would be about $0.045. I compared the standard Coca Cola can to the cheapest can by calculating the amount spent on 19 billion cans per year and what could be spent with the cheapest can.
Standard Coca Cola Can
Cost: $0.047 $0.047 • $19 billion = $893 million |
Cheapest Can
Cost: $0.045 $0.045 • $19 billion = $855 million |
The total number of savings would be $38 million if Coca Cola switched to the cheaper can instead of the continued use of the standard can. There some pros to this option, but there are also cons.
Pros
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Cons
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Reflection: This problem taught me a lot that may help me in future classes and maybe even in the long run. I have learned how to calculate surface area of a cylinder with only two variable, radius and volume. For my grade, I think I deserve a 9.5 out of 10 because I was very thorough with the problem by sticking to the instructions and investigating it even further. Two habits of a mathematician (HOAM) I used were Staying Organized and Conjecture & Test. I think I stayed organized by keeping all my calculations and work in one area where it corresponded with the information. In other words, I made sure that I had all my calculations organized to know what goes to what. I also tested a lot with calculating different radius' and heights for can #1 and can #2. In conclusion, I thought this was an interesting problem and it was a different way to look at volume and surface area.