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Math Help - Random Calculus Question

  1. #1
    Senior Member
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    Exclamation Random Calculus Question

    A 5 m ladder leans against a wall with a 1 m x 1 m crate between the ladder and the wall. a is the distance from the bottom of the wall to the base of the ladder where it is touching the ground. b is the distance from the bottom of the wall to the point the ladder is touching the wall.

    From the previous parts of the question, I've got the following answers:

    a + b = ab

     a^2 + b^2 = 25 (Pythagoras')

    (ab)^2 - 2ab - 25 = 0

    a = \frac{\sqrt{26} + 1}{b}

    b^4 - 25b^2 + 37.20 = 0

    From all the formulas I've found out, how do I find the maximum value of b?

    The answer's 4.84.

    I've tried using differentiating b^4 - 25b^2 + 37.20 = 0 and then equating it to 0 like similar questions but it doesn't seem to work...
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  2. #2
    Senior Member Twig's Avatar
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    Gothenburg
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    hi

    \frac{d}{db} (b^{4} - 25b^{2} +37.20) = 4b^{3} - 50b = 4b(b^{2} - \frac{50}{4}) = 0
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  3. #3
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    Exclamation

    That's what I did but the answer I got wasn't the same as the one in the solutions (which I've checked to be right).
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