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Math Help - lenght on a ladder

  1. #1
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    lenght on a ladder

    I have problem with this question :/

    a 2 meter high plank stands on the distanse 3 meter from a wall. Decide the shortest lenght a ladder must have to reach from the ground to the wall. (the wall and the plank is vertikal and the ground is horisontel.)

    the right answer is (3 + ^3(12^0.5) (1+(4 / 12^2/3))^0.5 but i dont get it :/

    i did this way:
    i say that the distance from the plank to the ladder is x

    L^2 = (3+x)^2 + (6/x + 2)^2
    2L`= 2(3+X) + 2(6/x + 2)(-6/x^2)
    L= 0 => x4 + 3x^3 - 12x - 36 = 0
    i dont know more, help please
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  2. #2
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    Quote Originally Posted by paulaa View Post
    I have problem with this question :/

    a 2 meter high plank stands on the distanse 3 meter from a wall. Decide the shortest lenght a ladder must have to reach from the ground to the wall. (the wall and the plank is vertikal and the ground is horisontel.)

    the right answer is (3 + ^3(12^0.5) (1+(4 / 12^2/3))^0.5 but i dont get it :/

    i did this way:
    i say that the distance from the plank to the ladder is x

    L^2 = (3+x)^2 + (6/x + 2)^2
    2L`= 2(3+X) + 2(6/x + 2)(-6/x^2)
    L= 0 => x4 + 3x^3 - 12x - 36 = 0
    i dont know more, help please
    This is incomprehensible, is there an associated diagram for this question.

    CB
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  3. #3
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    no just the information i gave :/
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by paulaa View Post
    no just the information i gave :/
    Then the question is either incomplete or the translation is confusing.

    CB
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  5. #5
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    I think this may be the set up ... goal is to minimize L
    Attached Thumbnails Attached Thumbnails lenght on a ladder-ladderprob.jpg  
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  6. #6
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    exacly
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  7. #7
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    I interpret this as meaning that the "2 meter high plank" forms a barrier three feet from the wall that the ladder must clear.

    Let x be the distance from the plank to the foot of the ladder, and y the height that the ladder reaches on the wall. Assuming the ladder just clears the barrier, we have, by "similar triangles", \frac{x}{2}= \frac{x+ 3}{h}.

    Essentially, then, you want to minimize (x+3)^2+ h^2 the length of the hypotenuse, subject to the condition that \frac{x}{2}= \frac{x+3}{h}. I don't see any good way to do that without Calculus.
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  8. #8
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    If angle of ladder with ground is 45 degrees L =5rad2=7.071 m.There is a miminum at approx 41 degrees
    L=7.023m

    bjh
    Last edited by bjhopper; March 16th 2011 at 04:37 PM. Reason: add answer
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  9. #9
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    A "ye olde" one:
    Google
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