A cylindrical can with a top is made to contain 1 liter of solutions. Find the dimensions that will minimize the cost of the metal to manufacture the can.

The volume is given as 1 liter which is equivalent to 1000ml or 1000cc (cubic centimeters)

The formula for the volume of a cylindrical can is `V=pi r^2h` where r is the radius of the top/bottom, and h the height of the can.

Letting V=1000 and solving for h in...

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The volume is given as 1 liter which is equivalent to 1000ml or 1000cc (cubic centimeters)

The formula for the volume of a cylindrical can is `V=pi r^2h` where r is the radius of the top/bottom, and h the height of the can.

Letting V=1000 and solving for h in terms of r we get `h=1000/(pi r^2)`

We want to minimize the amount of material used to form the can. Assuming that the sides are thin and of equal width, we want to minimize the surface area.

The formula for the surface area is `SA=2pir^2+2pirh` . Substituting for h we get `SA=2pir^2+2000/r`

To minimize the surface area function we take the first derivative and find the critical points:

`(dSA)/(dr)=4pir-2000/r^2` Setting the derivative equal to zero we get:

`4pir=2000/r^2`

`4pir^3=2000`

`r^3=500/pi`

`r=root(3)(500/pi)~~5.42`

So the radius we seek is approximately 5.42cm. Then the height is approximately 10.84 cm

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