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The Golden Ball & the Oxford Professors

  1. May 5, 2006 #1
    An excellent puzzle for you all - not overly hard, but a good tester :

    We have a ball made of pure gold. 2 Oxford Professor had obtained it in illicit fashion during an archaeological dig in Peru. The 2 accomplices being born complexifiers, fell into a dispute about how they should divide up this valuable object.

    One had a fancy to have a solid gold paperweight & as a ball is not much use for that purpose, decided that it must be a cylinder ; so he said,

    " All I want is a cylinder from the ball & I can turn this up on the lathe in the laboratory. All the rest of it, the golden swarf, you shall have & you can sell it for a considerable sum. "

    The 2nd Professor did some calculations & proved to his own satisfaction that any true cylinder from the sphere must contain less than half the volume of the sphere & so he agreed to the terms.

    Was he wise ?

    If the total weight of the of gold in the ball was 1 kg, what was the least weight of swarf his friend could make in turning a true cylinder from the golden ball ? "
     
  2. jcsd
  3. May 5, 2006 #2
    Is it 0.63kg?
     
  4. May 5, 2006 #3
    nope

    it's a high-school maths problem ( but probably last year of high school )
     
  5. May 5, 2006 #4

    matt grime

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    It's a high school maths problem for 14 year olds at most. It's a simple maths problem for people who've done maths to A-level equivalent.
     
  6. May 5, 2006 #5
    it involves some calculus ( to get a non-calculator answer ) - they didn't teach me that as a 14y old ( only when i was 17y & doing my A -level )
     
  7. May 5, 2006 #6

    berkeman

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    volume(cylinder) / volume(sphere) = 3 / 4*SQRT(2) = 0.53

    So the mass of the carved off part is 0.47kg

    Although I solved it as a square in a circle problem and extended that answer to 3-D. I need to think more about if that's fully valid....

    Happy Friday folks!


    (EDIT -- I just did it for the sphere, and get the same term in the differentiation, so same answer.)
     
    Last edited: May 5, 2006
  8. May 5, 2006 #7

    Curious3141

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    Something's wrong. I get [tex]max(\frac{V_{cyl}}{V_{sphere}}) = \frac{1}{\sqrt{3}}[/tex] which approximates 0.58. Second part is 0.42.
     
  9. May 5, 2006 #8
    correct !

    can you give us the detailed workings - it's nice for the public record :smile:
     
  10. May 5, 2006 #9

    Curious3141

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    Really ? I self-learned Calc (up to basic Integral Calc, including Volumes of Revolution) "for fun" when I was 11 or so. School taught (bored) me with it again at 15-16 before O levels.
     
  11. May 5, 2006 #10

    berkeman

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    Yeah, what'd I do wrong, I wonder. Did you get

    0 = d/dTheta ( sin(theta)cos(theta) ) ?
     
  12. May 5, 2006 #11

    Curious3141

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    Taking a section thru the middle of the sphere/cylinder it's obvious that to have any hope of maximising the cylinder's volume, you'll need to inscribe it symmetrically into the sphere (touching the inside).

    Then, in the cross-section, it just becomes a simple rectangle within a circle problem. Let the height of the cyl. be h Then the radius r of the cylinder is [tex]r = \sqrt{R^2 - \frac{h^2}{4}}[/tex]

    [tex]V_{cyl} = \pi r^2h = \pi(R^2 - \frac{h^2}{4})(h)[/tex]

    Differentiate that wrt h, set it to zero and solve for h, then find V_cyl, divide by 4/3*pi*R^3.
     
  13. May 5, 2006 #12

    Curious3141

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    I did it trigonometrically as well, if you're using the same theta I'm using, there should be a square on the sine term.

    More precisely, if [tex]\theta[/tex] is the angle subtended between a vertical line drawn from the center of the cylinder and a line drawn from the center of the cylinder to the point where the limiting disk of the cylinder touches the circumscribing sphere, then

    [tex]V_{cyl} = \pi R^2\sin^2\theta (2R\cos\theta) = 2\pi R^3\sin\theta \sin{2\theta}[/tex]

    max V_cyl when [tex]\tan{\theta} = \sqrt{2}[/tex]
     
    Last edited: May 5, 2006
  14. May 5, 2006 #13
    unfortunately for me, my library didn't stock maths primers then, i had to wait 'til i was 14y ole, won the local rotary club essay prize - tasted vol-a-vents for the 1st time, got a £15 book token - with which i bought my 1st maths book - boz & chaz

    more unfortunately, i didn't get a chance to do a maths degree after - i ended up as a cardiologist :frown:
     
  15. May 5, 2006 #14
    Could someone post the detailed workings from a - z , in "code" form - nice for the record ! :approve:
     
    Last edited: May 5, 2006
  16. May 5, 2006 #15

    Curious3141

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    You're a cardiologist ? I'm a Clinical Microbiologist ! Well, in training anyway. I've always felt I've missed my true calling by becoming a Physician.
     
  17. May 5, 2006 #16
    I assume you mean you missed your true calling by not becoming a Physician , as a Microbiologist isn't considered as having missed their true calling if they pass their MBBS or MD & then specialise in Microbiology

    A Clinical Microbiologist without a prior MBBS or MD , is not a Physician
     
  18. May 5, 2006 #17

    Curious3141

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    I have an MBBS. I am a Clinical Pathologist.:rolleyes:

    My greatest regret is not going to CalTech (I had admission and a scholarship). My parents wanted me to do Medicine locally.

    I consider my true calling to be within the Physical Sciences, pure or applied. I regret doing Medicine, which I consider to be intellectually unstimulating and a waste of time. Pursuing a non-clinical, academically-oriented discipline as a postgraduate is a compromise, making the best of a bad deal.
     
    Last edited: May 5, 2006
  19. May 5, 2006 #18
    More intriguing !?

    MBBS is a "English" degree

    They don't offer it to the Far East Guyz unless they study in England for at least 3y post-clinical...
     
  20. May 5, 2006 #19

    Curious3141

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    Please educate yourself on what Medical Degrees most former British colonies offer in their Universities. Singapore is a former Brit colony, FYI.
     
  21. May 6, 2006 #20

    matt grime

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    Ah, you wanted the exact amount cut off, rather than verifying that the professor in question was correct. I think there is a non-calculus method for doing it, but can't remember it. It's a famous problem, and certainly appears in the books of either Martin Gardner or David Wells.
     
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