If r is the radius the total area of the top and bottom discs is 2πr2, while the wrap-around part of the cylinder has a rectangular area of 2πrh, where h is the height of the can. The volume of the can is πr2h=500cc. Therefore, h=500/(πr2) cm.
The wrap-around part of the cylinder has an area of (2πr)(500/(πr2))=1000/r sq cm.
The cost in cents of the bottom and top discs is 6×2πr2=12πr2 cents, while the cost in cents of the wrap-around part is 4000/r cents, making the total cost C(r)=12πr2+4000/r cents.
When r=10cm, C(10)=1200π+400 cents, or 12π+4 dollars. If π=3.142, then C(10)=$41.70.