The Metal in Two Beverage Cans
Question: One Coke can is 15 centimeters long. The second coke can is 12 centimeters long. Both coke cans hold 355
milliliters of liquid.
What is the radius of the top of each coke can?
What is the volume of the surface area of the two can?
What can is better for the environment?
Answer: The procedure used to calculate the area of
metal on the surface of a coke can given the height of the can and the volume
of liquid in the cans is as follows.
Divide the volume of the can by the height of the can to get
the area of the top or cross section of a circle parallel to the top of the
can. (This follows from the observation
that the volume of the can is the product of the height and the circle around
the can.)
Get the radius of the circle around the can by solving for r
in the equation A=pi * r^{2}.
Get the area of the top and bottom of the cans. This is twice the area of the crosssection.
Get the area of the side of the can. This is 2 x pi x r x height.
Add the area of the top, bottom and side.
Calculations are laid out below.
Amount of Metal in Two
Coke Cans


Height of can in Centimeters

15

12

Volume of Can in
Milliliters

355

355

Area of cross section of
can in Centimeters

23.6667

29.5833

pi

3.1416

3.1416

Radius of crosssection
(Area/pi)^{0.5}
^{ }in centimeters

2.7447

3.0687

Area of Metal in Entire
Can (Sum of volume of top and bottom and side)
Square

306.0145

290.5381

Discussion of the
environmental significance of this result:
The standard can is 12 centimeters long but I have on my desk a can of La
Croix sparkling water that is 15 centimeters long. The can with a 15centimeter height has
around 5.33% more metal in it than a can with a 12centimeter height.
The typical empty soda can (one with a 12centimeter height)
weighs around 15 grams. This means the
can with a 15centimeter height weighs around 15.8 grams.
According to one web site the world consumes around 200
billion aluminum beverage cans each year.
Let’s assume that 5.0 percent of cans are nonstandard and weigh around
5.3% more than the standard. I am admittedly
uncertain about the validity of this assumption.
Based on this assumption I calculate tons of aluminum wasted
by assuming that 10 million cans deviate
from the standard can by having 5.3% more aluminum.
The extra aluminum stemming from the deviation for the
standard can is 7,990,188 grams or a bit over 8.8 tons of aluminum. (There are 907,185 grams in one U.S. ton.)
So if you want to somewhat reduce the amount of aluminum
used in the world and the amount of energy used to make this aluminum buy
355milliliter (12 ounce) soft drinks in cans that are 12 millimeters high
rather than soda in cans that is 15 millimeters high.
This leads to the question should government’s regulate the
size of soda cans? This question is
beyond the scope of the current post.
Note to math teachers: There are actually a lot more beverage can
math problems that should be created. You
could have your students calculate the area of metal given the radius of the
top. You could have them use Excel or calculus
to figure out the size of the can that uses the least amount of metal given the
volume of liquid in the can. More
problems of this type will follow.
What about the thickness and metal composition of each. Are they of equal portions?
ReplyDeleteWhat about the thickness and metal composition of each. Are they of equal portions?
ReplyDeleteOnly weight assumption used is 15 grams per standard can.
ReplyDelete