the question
What dimensions should a can have, in the conditions
to a given filling volume, noted with $V$, if a percentage $p$ of the material
used is lost during the making of the bottom and lid of the box because of
the shape of the cutout and the shape of the initial material, and the manufacturer wants to
use a minimum amount of material? (The shape of the box is a cylinder
right circular).
my idea
We know that the volume of a cylinder right circular is $V=h\pi R^2$, where h is its height and R is the radius of the base
Also, the material used to make the lateral surface is the same, without lost material, which means that it can be calculated as $h 2\pi R$.
The area of the lid and the bottom can be expressed as $2\pi R^2$, but we have to mention this after we eliminate that p percentage.
I don't know what to do next.
Best Answer
Use the AM-GM inequality to get $$2\pi hR + 2t\pi R^2 = \pi h R + \pi h R + 2t\pi R^2 \ge 3\pi(2th^2 R^4)^{1/3} = 3 \sqrt[3]{2\pi t V^2}$$ where $t = 1/(1-(p/100))$. Equality holds exactly when $hR = hR = 2tR^2$, ie $h = 2tR$, given $V \ne 0 \implies R \ne 0$.
At the minimum, you can plug in $h = 2tR$ into $V = \pi h R^2$ and solve to get $R = \sqrt[3]{V/2\pi t}$ and $h = \sqrt[3]{4t^2V/\pi}$.