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Thread: 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).

    $\displaystyle f(x) = axe^{bx}$



    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:

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

    Then I solved for a:

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

    I plugged that back into the original derivative, set it equal to 0, and solved for b, getting $\displaystyle \frac{-1}{6}$, 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; Oct 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).

    $\displaystyle f(x) = axe^{bx}$



    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 $\displaystyle -\frac{1}{b}$. That's all I've done, and I don't think it's right.
    First we know that

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

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

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

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

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

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

    $\displaystyle a=3e$

    So $\displaystyle f(x)=3exe^{-\frac{x}{6}}=3xe^{\frac{6-x}{6}}$
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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 $\displaystyle a$ and $\displaystyle b$ so that the function has a local maximum at the point (6, 18).

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


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

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

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

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


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

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

    . . 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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