An open cylindrical can of given capacity is to be made from a metal sheet of uniform thickness. If no allowance is to be made for waste material, what will be the most economical ratio of the radius to the height of the can ?

The curved surface of the cylindrical can be made by rolling the sheet and there will be no weste of material. But in order to have the base of the can, which will be circular of radius r we will have to cut it from a square metal sheet 2r by 2r and whose area will be $4r^2$ whereas the actual area of the base is only $\pi r^2$.But no allowance is to be made for waste of material. Hence the total surface area of the sheet used for making the open cylindrical can is $S=2\pi rh+4r^2$ 

Also $V=\pi r^{2}h(given)\therefore h=V/\pi r^2$

 

$\therefore S=2\pi r.\frac{V}{\pi r^2}+4r^2=\frac{2V}{r}+4r^2$

 

$\therefore \frac{ds}{dr}=-\frac{2V}{r^2}+8r=0or4r^3=V$ 

$\frac{d^{2}S}{d{r}^2}=\frac{4V}{r^3}+8=4.4+8=24+ive;$

hence minimum when $4r^3=V=\pi r^{2}h\therefore r/h=\pi /4$ is the required ratio.