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:Totalcost = [tex]\frac{24}{r}[/tex]+ [tex]\frac{2r^2}{3}[/tex]Step-by-step explanation:Given the volume of the can is 24 cubic cmAlso given that it costs 3 cents per square cm on the to and bottom sides And 2 cents per square cm on the curved sidesLet the radius and height of the can be r and hNow Volume =  π[tex]hr^{2}[/tex]24  = π[tex]hr^{2}[/tex]πh = [tex]\frac{24}{r^{2} }[/tex]Now for constructing we use the surface area which isTotal surface area =  lateral surface area + curved surface area Lateral suface area = 2π[tex]r^2[/tex] Cost for preparing the lateral surface is lateral surface / cost for top and bottom = [tex]\frac{2r^2}{3}[/tex]Curved surface area = 2πrh = 2r [tex]\times \frac{24}{r^2}[/tex] = [tex]\frac{48}{r}[/tex]Cost for preparing theCurved surface area is Curved surface / cost for curved  sides = [tex]\frac{48}{r\times 2}= \frac{24}{r}[/tex]Totalcost = [tex]\frac{24}{r}[/tex]+ [tex]\frac{2r^2}{3}[/tex]