A cylindrical can with height h and radius r is to be made to hold 1L (i.e. 1000cm3)of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
Formula for the volume of the cylinder is `V=r^2 pi h` which is equal to 1. From that we conclude that
`h=V/(r^2pi)=1/(r^2pi) ` (1)
Surface area of the cylinder (which we must minimize) is given by formula `S=2r^2pi+2rpih` ` `
Now we plug formula (1) into formula for surface area to get
To get minimum value of surface area we need to find minimum of function `S(r)=2r^2pi+2/r`
For that we first find where it's derivative is equal to 0.
` ` `4rpi-2/r^2=0`
Multiply by `r^2.`
Now we check to see that it is truly minimum by taking second derivation.
Since `S''(r)>0` (`4pi>0` and `4/r^3>0` because radius must be positive)we conclude that `r=root(3)(1/(2pi))` is indeed minimum value. If we plug that value of `r` into (1) we get `h=2root(3)(1/(2pi)).` i.e. height is twice as big as the radius.
Solution: `r=root(3)(1/(2pi)),` `h=2root(3)(1/(2pi))`