How do you construct a 12-ounce cylindrical can with the least amount of material?

1 Answer
Feb 23, 2015

I will do this only in metric units, but the idea will be clear.

Let's say we use #0.5L# as a target. This is equal to #500cm^3#

For the can (closed-topped) we need:
a top, a bottom and the outer surface.
Let's call #x# the radius (= half -diameter) of the can, and #y# the height

Then the volume will be:
#V=pix^2*y=500#
The total surface area #A# (=material):
Top+bottom: #pix^2# (times #2#)
Side surface: #2pix*y#
Total: #A=2pix^2+2pixy#

Since the volume is given, we can express #y# in #x#
#y=500/(pix^2)# and substitute

#A=2pix^2+2pix*(500/(pix^2))->#
#A=2pix^2+1000/x=2pix^2+1000x^-1#

Differentiating, we get #A'=2*2pix^1+(-1)1000x^-2#
#A'=4pix-1000/x^2# which we have to set to #0#

This leads to: #4pix*x^2=1000->x^3=1000/(4pi)->#
#x~~4.30# and diameter is double that.

#y=500/(pi*4.30^2)~~8.61#

Answer :
Diameter: #8.60cm# Height: #8.61cm#

Checking the answer: #V=pi*4.30^2*8.61=500.1...#
(close enough, with the rounding we've done)

Remark :
With the ounces and inches, you will have a lot of converting to do. I suggest you start by converting ounces to cubic inches.