Question

Calc1 Optimization:
A soup can is to be made in the shape of a cylinder, where the surface area of the label (the rectangle) is to be 100cm^2. Find the dimensions of the cylinder so that its volume is maximized, with the restriction that the radius r is 0 13 years ago

Answer 1

v = pir^2h
SA = 100cm^2
SA = r + h
100/(2pir) - r = h
I think this is what you need I haven't done an optimization problem in a while though. Do you know where to go from here?

Answer 2

sorry ignore the SA = r + h I was moving stuff around that should actually say SA = 2pir^2 + 2pirh

Answer 3

I substitute in the H into the volume formula, then multiply by the denominator of H, then simplify. afterwards, I set it equal to 0 to find the optimal radius by isolating r (since according to the other problems I did, the only variable left should be r)

Answer 4

Yes just think about it like this. When you are doing an optimization problem you are really solving for maxima and minima of a function. So you need to solve for the zeros of the function you are trying to find the optimal value for. Want to show me what you got?

Answer 5

I'll get back to you on that in a bit

Answer 6

Okay sounds good. Good luck.

Answer 7

sorry I mean the zeros if the derivative of the function you are trying to find the optimal value for** basically what you said you were going to do

Answer 8

of** not if haha it's late

Answer 9

no problem, thanks for really helping me out today. I'm curious how you got:
100/(2pir) - r = h

Answer 10

I factored 2pir out of the SA formula like this SA = 2pir(r+h) then I divided both side by 2pir then I subtracted r from both sides to get it in terms of H and I set the SA to the value you gave 100cm^2

Answer 11

ok i see it was just algebra. I'll keep working

Answer 12

Wait a second I just realized the SA is only for the rectangle not the full cylinder this changes things. It says the label only goes over the rectangle part.

Answer 13

how so?

Answer 14

The formula for SA I put is for the entire cylinder but your question only involves the rectangular part of the cylinder.

Answer 15

what changes then?

Answer 16

The formula for SA = 2pirh now so you have to sub h = 100/(2pir) instead I believe. Like I said I am rusty on this stuff tho.

Answer 17

Here's what I got:
SA = 2πrh
100 = 2πrh
100/(2πr) = h
50/πr = h

Volume = hπr^2
= πr^2 * (50/πr)
= r50
we hit a problem

Answer 18

yeah sorry I messed up earlier one sec I think I can fix it

Answer 19

(100-2pir^2)/(2pir) = h
v =2pir^2((100-2pir^2)/(2pir))
v = (200pir^2- 2pir^4)/(2pir)
never mind same issue as before I must be leading you in the wrong direction I am going to try and figure this out for a couple more minutes but I gotta go to sleep soon best of luck sorry I couldn't be more helpful.

Answer 20

I'll work on it some more. thanks for trying

Answer 21

Okay I think I got it after all I accidentally wrote v = 2pir^2h when I should have wrote v = pir^2h anyway
v = ((100pir^2-2pi^2r^4)/(2pir))
v = (50r-pir^3)
v' = 50 - 3pir^2
Now we find zeros

Answer 22

0 = 50 - 3pir^2
r = (50/(3pi))^(1/2)
sub that back into formula for volume without "h" and solve for volume
to get (50((50/(3pi))^(1/2))-pi((50/(3pi))^(1/2))^3)
V = 76.78
Now plug V back into formula this time with h and solve for h
76.78= pi(((50/(3pi))^(1/2))^2h
h = 76.78/(pi(((50/(3pi))^(1/2))^2)
h = 4.6
I think this is right, but as I am sure you could see I am pretty rusty with this. If you have the chance send me a message if it is right or wrong.