Question

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

Answer 1

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.