Question

1. A cylindrical can is to have a volume of 24p . The cost of the material used for the top & bottom of the can is 3 cents/ and the cost of the material used for the curved sides is 2 cents/. Express the cost of constructing the can as a function of its radius.

Answer 1

Totalcost =
`(24)/(r)`+
`(2r^2)/(3)`

Explanation:

Given the volume of the can is 24 cubic cm

Also given that it costs 3 cents per square cm on the to and bottom sides And 2 cents per square cm on the curved sides

Let the radius and height of the can be r and h

Now Volume = π
`hr^(2)`

24 = π
`hr^(2)`

πh =
`(24)/(r^(2) )`

Now for constructing we use the surface area which is

Total surface area = lateral surface area + curved surface area

Lateral suface area = 2π
`r^2`

Cost for preparing the lateral surface is lateral surface / cost for top and bottom =
`(2r^2)/(3)`

Curved surface area = 2πrh = 2r
`* (24)/(r^2)` =
`(48)/(r)`

Cost for preparing theCurved surface area is Curved surface / cost for curved sides =
`(48)/(r* 2)= (24)/(r)`

Totalcost =
`(24)/(r)`+
`(2r^2)/(3)`