# Why does a right-circular cylinder helps reduce surface area of the former International Prototype of Kilograms

I read on Wikipedia that right-circular cylinder shape helps reduce surface area of the former IPK, but could not find an explanation as to why. So how does such shape helps reduce its surface area? Wouldn't a spherical shape be better for that purpose?

• Hi, maybe it could help if you can tell where you read that
– Emmy
Sep 12, 2020 at 17:57
• @Emmy Thank you, I edited my post Sep 12, 2020 at 18:13
• Spheres are harder to make to precise tolerances, for one. You can't make a sphere by extrusion like you can with a cylinder, for example. Sep 12, 2020 at 18:49
• As far as I know the want to make a sphere , and can count the number of Si- atoms it is made of Sep 12, 2020 at 18:51
• @trula That's for Avogadro's number, and that sphere is made of silicon, rather than a platinum-iridium alloy (and is also extremely hard to make, requiring years of polishing by very skilled individuals who can "feel" imperfections in curvature to a tolerance better than a machine, most of whom are getting to be elderly these days). Silicon can't be extruded, and can be made to extremely high purity due to advances in the semiconductor industry, whereas the purity of the kilogram samples isn't as critical. It's a totally different situation. Sep 12, 2020 at 19:05

Though it's not precisely clear, it's possible that they meant something like this:

The surface area of a cylinder of unit volume is minimized when its height is equal to its diameter.

So it wasn't claiming that a cylinder was the minimum-surface-area shape, but rather that a cylinder of those particular dimensions had a lower surface area than cylinders of the same volume with different dimensions.

This statement can be proved fairly straightforwardly: a cylinder of diameter $$D$$ and height $$H$$ has volume $$V=\frac{1}{4}\pi D^2H$$ and surface area $$S=\frac{1}{2}\pi D^2+\pi DH$$. If we fix the volume to be a constant $$V_0$$, then we have that $$H=\frac{4V_0}{\pi D^2}$$. Substituting, we get the surface area as a function of the diameter:

$$S=\frac{1}{2}\pi D^2+\frac{4V_0}{D}$$

To minimize $$S$$, we first find points where the derivative with respect to $$D$$ is zero:

$$\frac{dS}{dD}=\pi D-\frac{4V_0}{D^2}=0$$

The solution to this is $$D=(4V_0/\pi)^{1/3}$$. To see whether this is a maximum or a minimum, we can find the second derivative at that point:

$$\frac{d^2S}{dD^2}=\pi+\frac{8V_0}{D^3}$$

This is always positive for $$D>0$$, so this is indeed a minimum.

So the diameter at which surface area is minimized is $$D=(4V_0/\pi)^{1/3}$$. This means that the height at which surface area is minimized, for a constant volume, is:

$$H=\frac{4V_0}{\pi(4V_0/\pi)^{2/3}}=(4V_0/\pi)^{1/3}=D$$