Question

Calculus help please: 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 surface area?

Answer 1

I am given both r and h, and am able to show that \[r=\sqrt[3]{V/(2 \pi)}\] , which is what is given

Answer 2

ick
make the diameter the height
i guess we have to do a bunch of work, but that is the punch line

Answer 3

V is fixed, it is a constant, treat it like one

Answer 4

what you need is a formula for the surface area

Answer 5

top and bottom have area \(\pi r^2\) for a total of \(2\pi r^2\)

Answer 6

h is given as \[h=2\sqrt[3]{V/(2 \pi)}\] , how do I find that with the r I have found? I keep coming up with other answers when I substitute r in either the volume or area formula for the tin can

Answer 7

area of the cylinder is \(2\pi rh\)

Answer 8

yes, plus top and bottom

Answer 9

total surface area is
\[2\pi r^2+2\pi h\] the latter from the circumference of the circle times the height

Answer 10

that is what you are trying to minimize

Answer 11

Volume is fixed at
\[V=\pi r^2h\]

Answer 12

solve that not for \(r\) but for \(h\) and get
\[h=\frac{V}{\pi r^2}\]

Answer 13

yes, I'm told to minimize the surface area \[2\pi r^2+2\pi rh\]

Answer 14

yes
replace \(h\) by \(\frac{V}{\pi r^2}\)

Answer 15

did that, took the derivative of the equation, using V as constant, set it to zero, and came up with the r= equation above

Answer 16

then you get a function of \(r\) namely
\[V(r)=2\pi r^2+\frac{V}{\pi r^2}\]

Answer 17

the derivative is
\[4\pi r-\frac{2V}{\pi r^3}\]

Answer 18

ooh i think i made a mistake sorry

Answer 19

so I have r (above) wondering why I keep getting a different h value than what the professor has given us as the correct answer (above)

Answer 20

\[V=2\pi r^2+2\pi rh\] right?

Answer 21

yes

Answer 22

put \(h=\frac{V}{2\pi r^2}\) and get
\[V(r)=2\pi r^2+\frac{2V}{r}\]

Answer 23

ok still screwing up, should be \(h=\frac{V}{\pi r^2}\) but \[V(r)=2\pi r^2+\frac{2V}{r}\] is right

Answer 24

not the way I went with it

Answer 25

taking derivatives gives
\[V'(r)=4\pi r-\frac{2V}{r^2}\] yes?

Answer 26

I took the derivative of the area fomula after solving the volume formula for h, and substituting that into the area formula

Answer 27

you are trying to minimize the surface area, not the volume

Answer 28

the volume is fixed, it is a number

Answer 29

@xapproachesinfinity was helping with this same problem last night, I got to the point where I found r, but was unable to show steps to find the given h (above)

Answer 30

oh i see you want r and h right?

Answer 31

yes, vol is fixed, need to minimize surface area, both r and h are given above for optimized surface area with volume V, I need to show the steps. I can get to the point of showing what r is, but then when I go to substitute that into either the volume or surface area formula, I keep getting a different answer than what the professor has provided.

Answer 32

seems you r is correct just did it
the h you did what i said yesterday correct and you didn't get a good formula matching what yu need

Answer 33

\[4\pi r^3-2V=0\\
4\pi r^3=2V\\
r^3=\frac{V}{2\pi}\\
r=\sqrt[3]{\frac{V}{2\pi}}\]

Answer 34

h is given above, can you show me how to find THAT h, using the r as satellite has providded?

Answer 35

correct satellite

Answer 36

solve
\[\pi r^2h=V\] for \(h\)

Answer 37

h=V/(pir^2)

Answer 38

using the r above

Answer 39

that's where I get screwed up, something I'm doing isn't working out, can you show me your algebra?

Answer 40

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

Answer 41

try this replace V by pi r^2h
cancel
and rewrite the radical
to h=2r

Answer 42

jeez this is hard to type

Answer 43

that is a good idea
again the punch line is the height is the diameter

Answer 44

why is it that the height is the diameter?

Answer 45

\[\pi(\sqrt[3]{\frac{V}{2\pi}})^2h=V\] we can use rational exponents i guess, maybe that will help

Answer 46

\[r=\sqrt[3]{\frac{\pi r^2h}{2\pi }}=\sqrt[3]{\frac{r^2h}{2}}\]

Answer 47

cube all sides to get \[r^3=\frac{r^2h}{2} \Longrightarrow h=2r\]

Answer 48

X, how did you know to look for that, experience, or looking to get rid of V?

Answer 49

\[h=V\times \left(\sqrt[3]{\frac{2\pi}{V}}\right)^2\frac{1}{\pi}\]

Answer 50

grind it out if you like, i like @xapproachesinfinity way more

Answer 51

yeah, tried grinding it out, and got a different answer than the prof...Satellite, any chance you'd grind it out so I can learn some algebra?

Answer 52

just after eyeballing, well i won't say i have enough experience :)
if you tried replacing r that is good too

Answer 53

guys, off to bed, thanks for all of the help, will check responses in the morning

Answer 54

\[h=V\times V^{-\frac{2}{3}}\] etc etc