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Math Help - Calculus - Maximum length of ladder

  1. #1
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    Calculus - Maximum length of ladder

    Attached is my working of the current problem, but i get an invalid solution. Any help would be appreciated
    Thanks
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  2. #2
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    Re: Calculus - Maximum length of ladder

    I'm not sure what happened on the last step: You have \frac{64\sin{x}}{\cos^2{x}}-\frac{48\cos{x}}{\sin^2{x}}=0, which should give you:

    \frac{48}{64}=\tan^3{x}, so x is approximately 0.7375 radians or 42.3 degrees.

    That gives you a ladder length of \frac{48}{\sin{x}}+\frac{64}{\cos{x}}=157.85.

    - Hollywood
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  3. #3
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    Re: Calculus - Maximum length of ladder

    thanks for your reply hollywood, Is it possible for you to exaplain how you got from what i had to what you have as i am new to calculus
    Thanks
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  4. #4
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    Re: Calculus - Maximum length of ladder

    Sure, but you already did the hard part...

    Starting with your equation (recall you took the derivative and set it equal to zero):

    \frac{64\sin{x}}{\cos^2{x}}-\frac{48\cos{x}}{\sin^2{x}}=0

    add \frac{48\cos{x}}{\sin^2{x}} to both sides:

    \frac{64\sin{x}}{\cos^2{x}}=\frac{48\cos{x}}{\sin^  2{x}}

    multiplly both sides by \frac{\sin^2{x}}{64\cos{x}} and combine the trig functions:

    \tan^3{x}=\frac{48}{64}

    Then take the cube root and inverse tangent to get x=0.7375. But you want the length of the ladder, so you have to plug in to the original equation:

    \text{length}=\frac{48}{\sin{0.7375}} + \frac{64}{\cos{0.7375}}=71.38+86.47=157.85.

    - Hollywood
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  5. #5
    Junior Member nimon's Avatar
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    Re: Calculus - Maximum length of ladder

    Interesting, I was just doing this problem in a book I'm reading. The solution I found doesn't use the angle at all, but instead labels the lengths with new variable names and finds a relationship between them in terms of the known variables. The answer is given in terms of a and b and I thought you might find it interesting. It's attached to my reply.
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