Create a can with a certain volume. The surface should be the lowest possible ensure a minimum amount of material. In this case I used the volume one liter as an example.
This task could also be useful for companies, because our ideal can would need the lowest possible amount of materials. Big companies can save millions of dollars by using this can.
Volume (cylinder): V=h*r²*π
Surface (cylinder): S=2r*π*h+2r²π
V=h*r²*π=1 =>h=1/(r²*π) (1) S(h,r)=2r*π*h+2r²*π (2) Insert (1) in (2): S(r)=(2r*π/r²*π)+2r²*π=2r²*π+(2/r) (3) S'(r)=4r*π-(2/r²) (4) 0=4r*π-(2/r²) =>2/r²=4r*π =>2=4r³*π =>1/2π=r³ =>r=(1/2π)^1/3=1,16245 (5) Insert (5) in (1): h=((2π)^2/3)/π=2^(2/3)*π^(-1/3)=1,08385 (6) both would be in dm
But what would be the solution, if the volume is not one liter:
V=h*r²*π =>h=V/(r²*π)=(insert (5)) V/((1/2π)^2/3)=0,74004*V =>r=(V/h*π)^1/2=(insert (6)) (V/(2π)^2/3)^1/2=0,54193*V^1/2