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Math Help - Nail in Can

  1. #1
    Junior Member BobBali's Avatar
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    Nail in Can

    Hi All,
    A similar question to the ladder one i posted earlier. Now that i have an idea from what i was shown earlier, i've tried to attempt it.
    The Question says: A 20cm nail fits inside a cylindrical can. Find the maximum radius of 3 balls that will fit exactly inside the can?

    I have drawn the diag to the best of my ability. At first i thought the nail stood upright in the can.
    So i the length of the can is 20cm and to fit 3 balls in =

    20/6 = radius = 3.33cm (wrong ans)

    But, using the help i got from the ladder question, I then thought the nail is probably lying diagonally (AB) and then i tried splitting the diag to make right angel triangles.

    *h = Radius

    (AO)^2 + h^2 = 10^2 > (AO)^2 + h^2 = 100

    (2AO)^2 + (2h)^2 = 20^2 > (4AO)^2 + (4h^2) = 400

    Subtracting the two doesn't work..
    I tried making (AO^2) the subject = AO^2 = 100 - h^2
    and sub it into the second equation...still nothing :-(
    Attached Thumbnails Attached Thumbnails Nail in Can-pic-2.png  
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  2. #2
    Super Member

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    Re: Nail in Can

    Hello, BobBali!

    Actually, this is not a max-min calculus problem at all!


    A 20 cm nail fits inside a cylindrical can.
    Find the maximum radius of 3 identical balls that will fit exactly inside the can?

    I have drawn the diagram to the best of my ability.

    I thought the nail is lying diagonally (AB) and I split the diag to make right triangles.

    *h = Radius

    \begin{array}{ccc}(AO)^2 + h^2 \:=\: 10^2 \\ \\ (2AO)^2 + (2h)^2 \:=\: 20^2 \end{array} . These two equations are equivalent!

    Consider the largest right triangle.

    The height is 2h.

    The base is 6h.

    The hypotenuse is 20.


    Therefore: . (6h)^2 + (2h)^2 \:=\:20^2 \quad\Rightarrow\quad 36h^2 + 4h^2 \:=\:400

    . . . . . . . . . 40h^2 \:=\:400 \quad\Rightarrow\quad h^2 \:=\:10 \quad\Rightarrow\quad h \:=\:\sqrt{10}

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  3. #3
    Junior Member BobBali's Avatar
    Joined
    May 2010
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    Tanzania
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    Re: Nail in Can

    Hi Soroban, Thank you! I substituted the above value of [tex]sqrt{10}=\pm 3.16[\math]

    into the equation constructed to cross check if i get the correct hypotenuse of 20cm and i obtained 19.98[tex]\simeq[\math] 20.

    However, in the textbook the ans is 3.03 and when that is substituted for h = 19.16 , which is a large margin of error, must be an error in the txtbook...
    Thank you ;-)
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