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Maximum and Minimum

  1. Jul 6, 2013 #1
    Can someone help me as a am racking my brain. I have been given this question;

    Two oil pipes with circular cross-sections are subject to the constraint that the sum of their
    diameters is 1m. By using careful reasoning find the diameters that give the maximum and
    minimum possible combined cross-sectional areas. In each case give the combined crosssectional
    area.

    Which I am unsure of. I dont get how you can get a maximum and minimum of a circle diameter when the volume of material going through it is irrelevant.

    Any assistance would be greatly appreciated.
     
  2. jcsd
  3. Jul 6, 2013 #2

    phinds

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    Can you not find a way to start attacking the problem mathematically? What is the area of a circle in terms of what you know?
     
  4. Jul 7, 2013 #3
    I believe I have cracked it, however will not know until it has been submitted so still need to try different angles to see what else I can come up with.
     
  5. Jul 8, 2013 #4

    verty

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    The question has nothing to do with angles. This is a basic optimization problem, your textbook will explain how to solve them.
     
  6. Jul 8, 2013 #5

    phinds

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    I take it you are not a native English speaker. "Angles" in this context has nothing to do with geometry, it mean "different ways to try something".
     
  7. Jul 13, 2013 #6
    So how did you work it out then? I can't seem to figure out how to solve your problem mathematically.
     
  8. Jul 13, 2013 #7
    You want to maximize [itex]\pi(r_a^2 + r_b^2)[/itex] based on the constraint that [itex]r_a + r_b = 1[/itex].
    Try starting with the identity [itex](r_a+r_b)^2 = r_a^2 + r_b^2 + 2r_ar_b[/itex].
     
  9. Jul 13, 2013 #8

    verty

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    This is a typical question that any calculus book will show you how to solve, usually under "maxima and minima" or "applications of the derivative". I would class it as an easy question if you know the method, which is why I said above that the poster should refer to his textbook for the method of solving these.

    We were all waiting for the poster to present some work because, if he knew the method, he would have been able to get started. If he got stuck somewhere, we could have helped. But being unable to start means he does not know enough to benefit from our help. And, it's against the rules.

    If the OP can show some work, I'm sure he will receive much help.
     
  10. Jul 13, 2013 #9

    haruspex

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    Let one of them have diameter x. What's the diameter of the other? What's the sum of the cross-sectional areas?
     
  11. Jul 13, 2013 #10

    verty

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    I think the question following this will be, what do I do next? And the only answer one can give to that is to explain in general how to solve these. I was hoping we could avoid that.

    Actually, I learned enough in school to solve this with just the most basic calculus knowledge. This isn't a big deal really, I shouldn't worry.
     
    Last edited: Jul 13, 2013
  12. Jul 13, 2013 #11
    It is solvable the way I gave without any general knowledge. Just substitute [itex]r_a = 1-r_b[/itex] or vice versa after arrangement.
     
  13. Jul 13, 2013 #12
    I started with the Area, substituted R for x and y. Worked out the starting area which was my maximum. Transposed the diameter formula to give me Y=1-X
    Differentiated the transposed area formula using the (1-X) which got rid of Y, so eventually I got x=0.5, I put this in the original area formula which gave me my minimum. I then done my graph with x=1 Y=0 (diameter) and went through x=.9 y=.1, x=.8 y=.2 and so on. This gave me my changing areas. This proved my answer and gave me a 'U' like graph.
     
  14. Jul 13, 2013 #13

    SteamKing

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    This is incorrect. According to the OP, the DIAMETERs must sum to 1 meter. Then, the sum of the areas of the two pipes should be expressed in terms of their diameters
     
  15. Jul 13, 2013 #14

    Ray Vickson

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    The problem as written has no solution, as you can have area =+infinity by taking radii of + and - infinity. Restricting radii to non-negative values gives a solvable problem, but not one to which calculus methods apply.
     
  16. Jul 13, 2013 #15

    verty

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    Pipes cannot have negative radii. At least no pipe I've ever seen. Perhaps a Klein bottle can, I don't know about that.
     
  17. Jul 13, 2013 #16

    Ray Vickson

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    Of course, and that is why your formulation was incomplete--it needed the sign restriction. Just because you and I know the variables must be >= 0 does not mean that a solver package would know it. If you were to submit your problem *exactly as written* to, eg., the Excel Solver it would essentially give you an unbounded solution. Sometimes we do not need to specify sign restrictions (eg., for the present case with *minimization*) but in other times we need to be more explicit.

    Anyway, in this example the maximum and minimum problems are completely different, even in the presence of sign-restricted radii. Calculus methods solve the min problem, but not the max problem.
     
    Last edited: Jul 13, 2013
  18. Jul 13, 2013 #17

    HallsofIvy

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    Or use [itex]r_1+ r_2= 1/2[/itex].
     
  19. Jul 13, 2013 #18
    Write an equation for the area in terms of one of the diameters eg in terms of d1. By inspecting the equation and bearing in mind that d1 can have a minimum value of zero and a maximum value of 1. simple reasoning, by reference to the equation, gives the maximum area and the minimum area .The latter value can be found by calculus. I guess most people here know the answers.
    To see it more clearly it may be helpful to sketch a graph of area against d1.
     
    Last edited: Jul 14, 2013
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