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Help with more optimization please, Maximum/minimum

  1. Jan 9, 2007 #1
    1. The problem statement, all variables and given/known data
    A cylindrical can is to hold 500 cm^3 of apple juice the design must take into account that the height must be between 6 and 15 cm, inclusive. How should the can be constructed so that a minimum amount of material will be used in the construction? assume no waste.

    2. Relevant equations
    V is volume and SA is surface area

    V= (pi)r^2h

    SA=2(pi)rh + 2(pi)r^2


    3. The attempt at a solution

    so far this is what I have:

    15>h>6

    v=(pi)r^2h
    500=(pi)r^2h
    500/(pi)r^2=h

    SA=2(pi)r(500/(pi)r^2) + 2(pi)r^2

    Now I believe I need to find the derivative and set it to zero then solve for r.

    The problem is that I can't quite figure out how I'm supposed to find the derivative if someone could give me a hand understanding the process I would really appreciate it as I have forgotten and my teacher is busy with the latest stuff.
    Thanks,
    ~RS
     
  2. jcsd
  3. Jan 9, 2007 #2
    i'm only 14, but i think i can help you. i did this problem before. it's been a long time, but i remember i got the answer of ratio 1:1 for the diameter to height of the can. this is independent of the volume, which you can give the constant V and it will cancel out when you attempt to minimize the ratio.
     
    Last edited: Jan 9, 2007
  4. Jan 9, 2007 #3
    S=2(pi)rh+2(pi)r^2. Replace h with V/((pi)r^2) so S is now a variable of just r (V is a constant) take derivative wrt to r and set to zero. solve for r. then calculate the ratio of r:h and the V will cancel out. you should get the answer 1:2. you still have to prove that the critical point gives a minimum but that is easy.

    << explicit solution deleted by berkeman >>

    this is ugly, but calculate the ratio r:h and V cancels out

    << explicit solution deleted by berkeman >>

    if you really want to be rigorous, calculate the second derivative of S and put r=(V/2p)^(1/3) into it and check that the result is positive, so that the extremum is indeed a minimum. that's quite messy so I'll leave you to do that.

    you said that the height must be between 6 and 15 cm, inclusive. to check that r=(V/2p)^(1/3) is consistent with this restriction, you will need to replace V with 500. If it isn't, then the one of the two endpoints will give you the minimum surface area.

    i find cylinders to be quite fascinating. for example, it can be twisted into a non-orientable manifold called the klein bottle. unlike the sphere and torus, it cannot be embedded in R^3, but only in R^4. i'm reading the proof of this but it is sooooooooooooo hard!
     
    Last edited: Jan 9, 2007
  5. Jan 9, 2007 #4

    berkeman

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    Staff: Mentor

    Tom1992,

    Do not post complete solutions for homework problems -- it is against the PF rules that you agreed to. Our task is to be tutorial, and to provide hints and guidance that helps the student figure out the answer.
     
  6. Jan 9, 2007 #5

    Gib Z

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    Homework Helper

    I think toms liking the attention, being 14 and knowing Calculus :) feels good doesn't it tom? But yea, dont post complete solutions. If you really want to know if your method/answer is correct as well, pm it to a mentor or someone smart. eg matt grime, mathwonk
     
  7. Jan 9, 2007 #6
    matt grime and mathwonk are mathematicians that run this forum?
     
  8. Jan 9, 2007 #7

    Gib Z

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    Homework Helper

    That do not run the forums, the Mentors and staff do. matt grime and mathwonk are mathematicians that are members of this forums, and they are very knowlegeable. They do not run it however
     
  9. Jan 9, 2007 #8
    By the way I dug up some old notes and got the method was confused by the pi, this question is now solved. Have to remember that you treat pi as a variable sometimes and as a number others.
     
  10. Jan 9, 2007 #9
    pi is a constant and is never treated as a variable because you don't solve for a value of pi (pi = 3.14159...) nor do you plot a function versus pi. i think what you mean is that pi often needs to be treated as a symbol that you manipulate in expressions as you do with x, and in that case yes.
     
  11. Jan 9, 2007 #10

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I was wondering if you meant you could not differentiate r2 and r! Glad to know that was not the case.

    But it can mean the function [itex]\pi(x)[/itex], the number of prime numbers less than or equal to x.
     
  12. Jan 9, 2007 #11
    true, but clearly he did not refer that any function in number theory
     
  13. Jan 10, 2007 #12

    Gib Z

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    Homework Helper

    [tex]\pi[/tex] can be used as a representative symbol for anything, let it be the number of prime numbers less than a variable, the ratio of the circumference to the diameter of a circle, or how many cakes various chefs can make. It just has to be explicitly stated in order to avoid confusion. Some uses have become standard notation, such as the Prime counting function, and the constant, but never the less there is nothing wrong with in.
     
  14. Jan 10, 2007 #13

    berkeman

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    Staff: Mentor

    Are you sure you're only 14? Dang!
     
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