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    Once more I have returned to the mathematical arena, and this time with a ground-breaking solution to the highly-technical and beautifully intricate problem which has hassled and beset academia to the highest degree for no less than four centuries.

    This morning in A2 Mathematics class our teacher (whom is of an obscure ethnicity - which I feel is a rather important note to add) presented his students with a supposed 'insanity-inducing' problem which up until that point in time I was completely unaware of. With a devilish smirk and lairy eyes he inquired - "If I wish to make a cylindrical tin can of volume 1000cm3, what is the minimum amount of tin in cm2 that I would require to achieve just this?" I was rapt with wonder and awe. I began scrawling and slowly but surely developing ever more complex equations in my make-shift notepad - unwittingly carving and creating completely novel fields of mathematics as I went - until the solution become clear. With this answer to the problem in mind I then jolted in raw excitement as I considered the possibilities of not merely working with a numerical volume, but considering a generalised volume 'V' and then seeing in the general case what kind of dimensions would be required for the radius and height of the tin can in order to minimize tin usage!

    It was brilliant, and completely unexplored territory. I declared these musings to my teacher, still in shock at the idea that I managed to solve his fiendish problem - to which he cried "IMPOSSIBLE" whilst slamming his hands fiercely on the desk. I was not daunted in the slightest; I stood my ground and completely disregarded his pessimism - instead I darted out of the room and back to my lair to consider the problem in question in its deepest and most elusive phases of generality.

    I bring to you now my proof of, and solution to, the 'Bean Can Conundrum'. Please enjoy folks. I must add as a minor disclaimer that this proof brings together a wide range of esoteric mathematical techniques from a variety of complex fields - incorporating ideas from mind-boggling areas such as Advanced Group Theory, Quantum Topology, Differential Geometry, Agricultural Development, Horticulture and Global Economic Theory III. If you feel the proof progressing at too challenging a pace then please feel free to skip to the final result at the end of this paper.

    We begin with a cylinder, with volume 'V', base radius 'r' and height 'h' such that V = \pi {r^2}h which implies that h = \frac{V}{{\pi {r^2}}}

    Now we note that the surface area, S, is given by S = 2\pi {r^2} + 2\pi rh

    Thus... S = 2\pi {r^2} + 2\pi r\left( {\frac{V}{{\pi {r^2}}}} \right) (Don't say I did not warn you about the complexity)

    S = 2\pi {r^2} + \frac{{2V}}{r}

    Differentiating and setting equal to 0...

    \begin{array}{l}

\frac{{dS}}{{dr}} = 4\pi r - \frac{{2V}}{{{r^2}}}\\

{r^3} = \frac{V}{{2\pi }}

\end{array}

    Which obviously implies that {r^3} = \frac{{\pi {r^2}h}}{{2\pi }} = \frac{{{r^2}h}}{2}

    Which in conclusion means that (brace yourself young grasshopper!) 2r = h

    DO YOU KNOW WHAT THIS MEANS? CAN YOU COMPREHEND THE SHATTERING REPERCUSSIONS OF SUCH A STATEMENT?

    It means that if you want to make a cylinder or tin or bean can which can hold anything up to a volume 'V', THEN TO MINIMIZE THE AMOUNT OF TIN YOU NEED TO MAKE THIS CAN THEN ALL YOU HAVE TO DO IS MAKE THE DIAMETER OF THE TIN CAN THE SAME LENGTH AS THE HEIGHT OF THE TIN CAN.


    The amateurs at Heinz haven't a clue - just look at the mismatch in dimensions here! THE LUDICROUS AND INTOLERABLE BUFFOONS.

    I have hereby revolutionized the bean can market, and I would now like to take the opportunity thank you all for your kind and continued support of my many ingenious ideas and theorems. I gladly accept (for the third consecutive year) my Fields Medal nomination - and my nomination to captain the Papua New Guinea IMO team; I am truly honoured beyond words.

    Until the next time my dear friends I bid you all farewell. :pierre:
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    At first I disregarded your thesis on the basis that no such fundamental an insight could have slipped past civilisation's giants for so long. Hah! How naïve, howparochial my former worldview! How dank and claustrophobic the confines of society's collective delusion seem from the pedestal of enlightenment! Reading http://www.cancentral.com/standard.cfm sickens me, to have created the entire tin mining and smelting industry around excessive usage and the pretence that there is no minimum surface area for constant volume: Heinz is completely bereft of corporate ethics. No doubt CEO William R Johnson makes a sizeable profit from protection money extorted from the tin industry, threatening to spill the beans to the mathematical elite about the fact that a large part of tin production is superfluous.

    I urge you all to write to Mr Johnson in the strongest possible terms to respect the natural rights of man by reversing his perennial, pernicious and oh-so-effective lie.




    P.S.
    I don't think the Fields medal committee is qualified to judge your work.
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    Crack is an incredible drug.
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    Excellent analysis, however:

    Let's say the pictured can has volume V,

    if I were to reduce the height of the can so that I may increase the diameter until it is the optimum (height = diameter) will it not be difficult to grasp the can within my palm for volume V of the can?

    So there are practical reasons.
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    (Original post by Benjy100)
    Once more I have returned to the mathematical arena, and this time with a ground-breaking solution to the highly-technical and beautifully intricate problem which has hassled and beset academia to the highest degree for no less than four centuries.

    This morning in A2 Mathematics class our teacher (whom is of an obscure ethnicity - which I feel is a rather important note to add) presented his students with a supposed 'insanity-inducing' problem which up until that point in time I was completely unaware of. With a devilish smirk and lairy eyes he inquired - "If I wish to make a cylindrical tin can of volume 1000cm3, what is the minimum amount of tin in cm2 that I would require to achieve just this?" I was rapt with wonder and awe. I began scrawling and slowly but surely developing ever more complex equations in my make-shift notepad - unwittingly carving and creating completely novel fields of mathematics as I went - until the solution become clear. With this answer to the problem in mind I then jolted in raw excitement as I considered the possibilities of not merely working with a numerical volume, but considering a generalised volume 'V' and then seeing in the general case what kind of dimensions would be required for the radius and height of the tin can in order to minimize tin usage!

    It was brilliant, and completely unexplored territory. I declared these musings to my teacher, still in shock at the idea that I managed to solve his fiendish problem - to which he cried "IMPOSSIBLE" whilst slamming his hands fiercely on the desk. I was not daunted in the slightest; I stood my ground and completely disregarded his pessimism - instead I darted out of the room and back to my lair to consider the problem in question in its deepest and most elusive phases of generality.

    I bring to you now my proof of, and solution to, the 'Bean Can Conundrum'. Please enjoy folks. I must add as a minor disclaimer that this proof brings together a wide range of esoteric mathematical techniques from a variety of complex fields - incorporating ideas from mind-boggling areas such as Advanced Group Theory, Quantum Topology, Differential Geometry, Agricultural Development, Horticulture and Global Economic Theory III. If you feel the proof progressing at too challenging a pace then please feel free to skip to the final result at the end of this paper.

    We begin with a cylinder, with volume 'V', base radius 'r' and height 'h' such that V = \pi {r^2}h which implies that h = \frac{V}{{\pi {r^2}}}

    Now we note that the surface area, S, is given by S = 2\pi {r^2} + 2\pi rh

    Thus... S = 2\pi {r^2} + 2\pi r\left( {\frac{V}{{\pi {r^2}}}} \right) (Don't say I did not warn you about the complexity)

    S = 2\pi {r^2} + \frac{{2V}}{r}

    Differentiating and setting equal to 0...

    \begin{array}{l}

\frac{{dS}}{{dr}} = 4\pi r - \frac{{2V}}{{{r^2}}}\\

{r^3} = \frac{V}{{2\pi }}

\end{array}

    Which obviously implies that {r^3} = \frac{{\pi {r^2}h}}{{2\pi }} = \frac{{{r^2}h}}{2}

    Which in conclusion means that (brace yourself young grasshopper!) 2r = h

    DO YOU KNOW WHAT THIS MEANS? CAN YOU COMPREHEND THE SHATTERING REPERCUSSIONS OF SUCH A STATEMENT?

    It means that if you want to make a cylinder or tin or bean can which can hold anything up to a volume 'V', THEN TO MINIMIZE THE AMOUNT OF TIN YOU NEED TO MAKE THIS CAN THEN ALL YOU HAVE TO DO IS MAKE THE DIAMETER OF THE TIN CAN THE SAME LENGTH AS THE HEIGHT OF THE TIN CAN.


    The amateurs at Heinz haven't a clue - just look at the mismatch in dimensions here! THE LUDICROUS AND INTOLERABLE BUFFOONS.

    I have hereby revolutionized the bean can market, and I would now like to take the opportunity thank you all for your kind and continued support of my many ingenious ideas and theorems. I gladly accept (for the third consecutive year) my Fields Medal nomination - and my nomination to captain the Papua New Guinea IMO team; I am truly honoured beyond words.

    Until the next time my dear friends I bid you all farewell. :pierre:
    You absolute legend. How did I not find out about this?!

    P.s on a serious note I suppose there are practical reasons why Heinz makes their tins thin and tall...it's easier to fit in the fridge
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    Um, you forgot to differentiate again to make sure it's a minimum
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    (Original post by shamika)
    Um, you forgot to differentiate again to make sure it's a minimum
    You don't need to. With constant volume, let the radius tend to 0, and the surface area tends to infinity: evidently larger than the stationary case's surface area calculated earlier.
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    Why even bother? You could have modelled the entire problem in 2d, for example minimising perimeter, maximising area with a rectangle. You said it yourself - have one pair of sides the same length as the other pair of sides

    ("MAKE THE DIAMETER OF THE TIN CAN THE SAME LENGTH AS THE HEIGHT OF THE TIN CAN.")

    That's just a square.
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    (Original post by Benjy100)
    ...
    PRSOM
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    (Original post by henpen)
    You don't need to. With constant volume, let the radius tend to 0, and the surface area tends to infinity: evidently larger than the stationary case's surface area calculated earlier.
    True (my point was more make the point explicit rather than how it is done )
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    (Original post by Benjy100)
    ...
    Zounds! This man is a marvel.

    Is there no stopping him?

    Has Heinz been squashed?

    Does Beanz Meanz Benjys?

    Can nothing be done?


    Watch out for next week's enthralling episode on the benjybatchannel.
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    ..
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    Yes we can
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    I'm sure I've seen a similar thread before?
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    LOL a 'new' proof / theory and it gets posted on the TSR first. Genius.
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    Truly astonishing!
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    Amazingly mind-boggling that it took us 4.5 billion years to find out !!!
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    They asked you how to minimize the amount of material used ?
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    I'm sure I've seen this thread at least twice before...
 
 
 
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