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Math Help - The problem of the ladder!

  1. #1
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    The problem of the ladder!

    I hope you can help me on to the next problem!

    I need to enter a ladder to a tower (not considered wide). What is the minimum height of the door (h) that makes this possible?




    How I can solve this problem?

    Thanks!!
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  2. #2
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    Quote Originally Posted by uruleo View Post
    I hope you can help me on to the next problem!

    I need to enter a ladder to a tower (not considered wide). What is the minimum height of the door (h) that makes this possible?




    How I can solve this problem?

    Thanks!!
    1. I've drawn a sketch of the worst case. (see attachment)

    2. You get 2 similar triangles. Use proportions:

    \dfrac hy = \dfrac x{27}~\implies~y = \dfrac{27 h}{x}

    3. Use the Pythagorean theorem:

    (h+x)^2+(27+y)^2=125^2

    4. Replace y by the term of #2 and solve for h:

    h(x)=\dfrac{x(125-\sqrt{x^2+729})}{\sqrt{x^2+729}}

    5. Differentiate h wrt x. Solve \frac{dh}{dx} = 0 for x. You should come out with x = 36. Which means that the minimum value of h = 64.
    Spoiler:
    for confirmation only: h'(x)= \dfrac{91125-\sqrt{(x^2+729)^3}}{\sqrt{(x^2+729)^3}}
    Attached Files Attached Files
    Last edited by earboth; April 28th 2011 at 11:46 PM.
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  3. #3
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    Thanks!!

    But i did not understand the step 4..
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  4. #4
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    Quote Originally Posted by uruleo View Post
    Thanks!!

    But i did not understand the step 4..
    I'm not sure what you mean ...(?)

    1. You have y = \dfrac{27h}x and (h+x)^2+(27+y)^2=125^2 which becomes:

    (h+x)^2+\left(27 + \frac{27h}x\right)^2=125^2

    (h+x)^2+\left(\frac{27}{x}\right)^2 (x+h)^2=125^2

    (h+x)^2=\dfrac{125^2x^2}{x^2+27^2}

    2. Can you take it from here?
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  5. #5
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    I think I got it, but I would appreciate if you specify each of the steps.
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