# Local Extrema Question

• Oct 26th 2009, 08:37 PM
lysserloo
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!
• Oct 26th 2009, 09:01 PM
TheEmptySet
Quote:

Originally Posted by lysserloo
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 $\displaystyle f(x)=3exe^{-\frac{x}{6}}=3xe^{\frac{6-x}{6}}$
• Oct 26th 2009, 09:04 PM
lysserloo
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.
• Oct 26th 2009, 09:16 PM
Soroban
[size=3]Hello, lysserloo!['/size]

Quote:

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

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

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?

• Oct 26th 2009, 10:03 PM
lysserloo
Oh wow, that explanation was PERFECT. Thank you SO much, it's much more clear now!