Find the minimum amount of tin sheet that can be made into a closed cylinder havin g a volume of 108 cu. Inches in square inches.

Answer 1

h = 108/r^2
a = 2pir^2 + 2pi r h
= 2pi r^2 + 2pi r (108/r^2)
= 2pi r^2 + 216pi/r

da/dr = 4pi r - 216pi/r^2
= 4pir^2(r^3-54)

da/dr = 0 when r = 3∛2

a(3∛2) = 2pi(9∛4) + 216pi/3∛2
= (2pi(9∛4)(∛4) + 216pi)/6
= 6pi∛4 + 36pi
= 6pi(∛4+6)

Answer 2

oops in the first line, Steve forgot the π

h = 108/(πr^2)

then

a = 2πr^2 + 216/r
da/dr = 4πr - 216/r^2
= 0 for a min of a

πr = 54/r^2
r^3 = 54/π
r = ∛(54/π) = appr 2.58
then h= 108/(πr^2) = 5.16

notice diameter = 5.16 and height = 5.16

Answer 3

good catch, Reiny.

Answer 4

wrong