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Math Help - Maximization

  1. #1
    Forum Admin topsquark's Avatar
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    Maximization

    Here's a problem that's been bugging me. A student asked me this a couple of days ago and I just can't get out of the circles it's twisting.

    (See diagram below.)
    A 15 ft ladder is constrained to keep one point on the ground. The ladder is propped up by a vertical 8 ft support that can be moved back and forth. What is the maximum horizontal extension of the ladder? (Marked as the line y in the diagram.)
    I can't seem to find a way to reduce the possible equations to something with one dependent variable. (This is supposed to be a Calc I problem.) Any takers?

    -Dan
    Attached Thumbnails Attached Thumbnails Maximization-maximization.jpg  
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  2. #2
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    The length of ladder between the ground and the support point is 8/sin(θ). The length of ladder above the support point is therefore 15 – 8/sin(θ), and the horizontal projection is y = (15 – 8/sin(θ))cos(θ).

    The maximum occurs when \sin\theta = \frac2{15^{1/3}}, at which point y = (15^{2/3}-4)^{3/2}\approx 3.00458.
    Last edited by Opalg; April 13th 2008 at 04:31 AM. Reason: Added numerical result
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  3. #3
    Moo
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    Hello,

    Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?
    Nice spot, Moo. And the answer is consistent with the one found using Oplag's suggestion.


    Edit: Which is now obvious following Opalg's edit lol!
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?
    That's likely to be because the other poster might be the student I referred to here.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Opalg View Post
    The length of ladder between the ground and the support point is 8/sin(θ). The length of ladder above the support point is therefore 15 – 8/sin(θ), and the horizontal projection is y = (15 – 8/sin(θ))cos(θ).

    The maximum occurs when \sin\theta = \frac2{15^{1/3}}, at which point y = (15^{2/3}-4)^{3/2}\approx 3.00458.
    Dang it all! I should have seen that. Nice work.

    -Dan
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