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

Feb 23, 2015

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

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

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:
$V = \pi {x}^{2} \cdot y = 500$
The total surface area $A$ (=material):
Top+bottom: $\pi {x}^{2}$ (times $2$)
Side surface: $2 \pi x \cdot y$
Total: $A = 2 \pi {x}^{2} + 2 \pi x y$

Since the volume is given, we can express $y$ in $x$
$y = \frac{500}{\pi {x}^{2}}$ and substitute

$A = 2 \pi {x}^{2} + 2 \pi x \cdot \left(\frac{500}{\pi {x}^{2}}\right) \to$
$A = 2 \pi {x}^{2} + \frac{1000}{x} = 2 \pi {x}^{2} + 1000 {x}^{-} 1$

Differentiating, we get $A ' = 2 \cdot 2 \pi {x}^{1} + \left(- 1\right) 1000 {x}^{-} 2$
$A ' = 4 \pi x - \frac{1000}{x} ^ 2$ which we have to set to $0$

This leads to: $4 \pi x \cdot {x}^{2} = 1000 \to {x}^{3} = \frac{1000}{4 \pi} \to$
$x \approx 4.30$ and diameter is double that.

$y = \frac{500}{\pi \cdot {4.30}^{2}} \approx 8.61$

Diameter: $8.60 c m$ Height: $8.61 c m$
Checking the answer: $V = \pi \cdot {4.30}^{2} \cdot 8.61 = 500.1 \ldots$