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Math Help - High school calculus problem. Is the question dodgy or am I missing something?

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    High school calculus problem. Is the question dodgy or am I missing something?

    Hi guys.

    So I've all up to (c), but in (d) the question talks about how x can be between 0 to 11 although in the context x must be less than 10 since the length of the wire is 10. This does not make sense to me. Or am I missing something?

    Please let me know

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    Attached Thumbnails Attached Thumbnails High school calculus problem. Is the question dodgy or am I missing something?-1618168_711861105551882_3892447209235541766_o.jpg  
    Last edited by willy0625; August 20th 2014 at 10:27 PM.
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    It would be a typo, it should be [0, 10].
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    If it was indeed a typo, would there be a maximum? I don't think there is because x is in (0,10), since (d) says two square are formed.
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    If it's possible to have an area from a finite amount, then it must have a maximum and a minimum.

    Maxima and minima can occur either at stationary points or endpoints.
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    Were you able to do parts a, b, and c? If so, what answers did you get?
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    a. I could show this

    b. \dfrac{dA}{dx}=\dfrac{1}{8}(2x-10)=\dfrac{1}{4}(x-5)

    c.  x= 5 since \dfrac{d^{2}A}{dx^{2}}(x=5)=-\dfrac{5}{4}<0.
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    So what now with question d?
    I don't think it is possible to find a maximum that fits the context, although, it actually does happen and x=0 and x=10.
    What now?
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    substitue them
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    Re: High school calculus problem. Is the question dodgy or am I missing something?

    Quote Originally Posted by willy0625 View Post
    a. I could show this

    b. \dfrac{dA}{dx}=\dfrac{1}{8}(2x-10)=\dfrac{1}{4}(x-5)

    c.  x= 5 since \dfrac{d^{2}A}{dx^{2}}(x=5)=-\dfrac{5}{4}<0.
    Check your second derivative.
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