Create a can with a certain volume.The surface should be lowest possible.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*r^2*π -Surface(cylinder): S=2r*π *h+2r^2*π
V=h*r²*π=1 =>h=1/r²*π (1) S(h,r)=2r*π*h+2r²π (2) Insert (1) in (2) S(r)=(2r*π/r²*π)+2r²*π=(2/r)+2r²*π (3) S'(r)=4r*π-(2/r²) (4) 0=4r*π-(2/r²) =>2/r²=4r*π =>2=4r³*π =>1/2π=r³ =>(1/2π)^1/3=r