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Math Help - Local Extrema Question

  1. #1
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    Local Extrema Question

    Problem: Find values of a and b so that the function has a local maximum at the point (6, 18).

    f(x) = axe^{bx}



    How do I solve this? Could you show your steps so I can see what you're doing and why? Also, how do you solve it to be a local maximum or minimum? I don't understand the concept of this problem at all, much less this specific problem.


    EDIT: I found the value of b by taking the derivative of the original equation, which is:

    f\prime(x) = ae^{bx}(1 + bx)

    Then I solved for a:

    a = \frac{18}{e^{6b}(1 + 6b)}

    I plugged that back into the original derivative, set it equal to 0, and solved for b, getting \frac{-1}{6}, which is correct.

    However, I can't figure out how to solve for a now! I keep getting an answer of 0, which I know isn't right!
    Last edited by lysserloo; October 26th 2009 at 09:00 PM.
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  2. #2
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    Quote Originally Posted by lysserloo View Post
    Problem: Find values of a and b so that the function has a local maximum at the point (6, 18).

    f(x) = axe^{bx}



    How do I solve this? Could you show your steps so I can see what you're doing and why?

    So far, I've taken the derivative and found the critical point, which I think is -\frac{1}{b}. That's all I've done, and I don't think it's right.
    First we know that

    18=6ae^{6b} \iff a=3e^{-6b}

    Now if we take the derivative we know that when x=6,f'(x)=0 becuase it is a maximum.

    f'(x)=ae^{bx}(1+bx)
    so we have

    f'(6)=0=ae^{6b}(1+6b)

    So one of the two above factors must be zero. We will choose the 2nd one (why?)

    1+6b=0\iff b=-\frac{1}{6} from above now we know that

    a=3e

    So f(x)=3exe^{-\frac{x}{6}}=3xe^{\frac{6-x}{6}}
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  3. #3
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    I've gotten as far as solving for b; I just edited my original post.

    How did you get 3e for a? That's the only part I'm stuck on now. Solving for a.

    EDIT: Ope! I understand! Never mind.
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  4. #4
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    [size=3]Hello, lysserloo!['/size]

    Find values of a and b so that the function has a local maximum at the point (6, 18).

    . . f(x) = axe^{bx}
    Your work is correct!


    The derivative is: . f'(x) \:=\:ae^{bx}(bx+1)
    . . So the critical value is: . x \:=\:-\frac{1}{b}

    But we are told that the maximum is at (6, 18) . . . That is, x = 6

    . . So we have: . -\frac{1}{b} \:=\:6 \quad\Rightarrow\quad b \:=\:-\frac{1}{6}

    The function (so far) is: . f(x) \;=\;axe^{-\frac{x}{6}}


    The point (6, 18) tells us that: . f(6) = 18

    So we have: . a\!\cdot\!6\!\cdot\! e^{-1} \:=\:18  \quad\Rightarrow\quad a \:=\:3e

    . . Got it?

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  5. #5
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    Oh wow, that explanation was PERFECT. Thank you SO much, it's much more clear now!
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