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.

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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...

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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

`S=2r^2pi+2rpi*1/(r^2pi)=2r^2pi+2/r`

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.

`S'(r)=0`

` ` `4rpi-2/r^2=0`

Multiply by `r^2.`  

`4r^3pi-2=0`

`r=root(3)(1/(2pi))` 

Now we check to see that it is truly minimum by taking second derivation.

`S''(r)=4pi+4/r^3`

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))`   

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