Question

A tin can (right circular cylinder) with top and bottom is to have volume V . What dimensions (the radius of the bottom and the height) give the minimum total surfacearea?

Answer 1

I get \[r=\sqrt[3]{V/(2 \pi)}\], that sound right?

Answer 2

this is calculus question correct

Answer 3

correct

Answer 4

we are minimizing
\[A=2\pi r^2+2\pi r h\]

Answer 5

actually your question is vague somewhat lol

Answer 6

yes, so I used the folume formula to solve for h, and substituted into the formula above to put in terms of r only, took that derivative with respect to r, and came up with the above

Answer 7

so, I understand V to be a constant

Answer 8

you are looking for two dimensions that minimize the surfacearea
not just r

Answer 9

yes, and they will be in terms of V

Answer 10

\[V=h(\pi)r^2 , A=h2(\pi)r + 2(\pi)r^2\]

Answer 11

well write h in terms of v and r
so you have only one variable for the A

Answer 12

\[A=2V/r+2 \pi r^2\]

Answer 13

do the derivative

Answer 14

solve for A'=0

Answer 15

\[r=\sqrt[3]{V/(2 \pi)}\]

Answer 16

alright if that's the answer should be good

Answer 17

which equation would I best use to solve to find h

Answer 18

the Volume equation

Answer 19

any chance you'd walk me through your steps in doing so?

Answer 20

once r is found h is found easily
that's what that equation is doing

Answer 21

you already did my friend you don't have to do anything else

Answer 22

what do you get for h?

Answer 23

you just use v=pir^2h to find h
so you have two equation one for r and one for h

Answer 24

i didn't do it
you just use \[V=\pi r^2h \Longrightarrow r=\sqrt{V/\pi h}\]
you replace r with this

Answer 25

and solve for h

Answer 26

r is the third root of RHS

Answer 27

\[\sqrt{V/\pi h}=\sqrt[3]{V/2\pi} \Longrightarrow V/h\pi=(V/2\pi)^{2/3 }\]

Answer 28

continue with this

Answer 29

the goal is to find two formulas
one for r and one for h
and that is all

Answer 30

thanks for the help X

Answer 31

no problem :)