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

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

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Answer 1 · 2015-02-23T00:45:17

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

Let's say we use $0.5 L$ $0.5L$ as a target. This is equal to $500 c m^{3}$ $500cm^3$

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

Then the volume will be:
$V = π x^{2} \cdot y = 500$ $V=pix^2*y=500$
The total surface area $A$ $A$ (=material):
Top+bottom: $π x^{2}$ $pix^2$ (times $2$ $2$ )
Side surface: $2 π x \cdot y$ $2pix*y$
Total: $A = 2 π x^{2} + 2 π x y$ $A=2pix^2+2pixy$

Since the volume is given, we can express $y$ $y$ in $x$ $x$
$y = \frac{500}{π x^{2}}$ $y=500/(pix^2)$ and substitute

$A = 2 π x^{2} + 2 π x \cdot (\frac{500}{π x^{2}}) \to$ $A=2pix^2+2pix*(500/(pix^2))->$
$A = 2 π x^{2} + \frac{1000}{x} = 2 π x^{2} + 1000 x^{- 1}$ $A=2pix^2+1000/x=2pix^2+1000x^-1$

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.

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

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