families of curves --- help!

**Paige05** · November 20th 2007, 01:17 PM

heres the question:

find a formula for a function of the form y = (ax^b)(lnx) where a and b are nonzero constants, which has a local maximum at the point (e^2,6e^-1)

i tried to create 2 equations and set and solve for the variables, but its giving me a rediculous answer so i dont think its right.

anyone know what to do?

**TKHunny** · November 20th 2007, 07:14 PM

Sadly, you did not show your equations or your solution.

You are right, though. The answers are not all that complicated.

**topsquark** · November 21st 2007, 08:16 AM

Originally Posted by Paige05

heres the question:

find a formula for a function of the form y = (ax^b)(lnx) where a and b are nonzero constants, which has a local maximum at the point (e^2,6e^-1)

i tried to create 2 equations and set and solve for the variables, but its giving me a rediculous answer so i dont think its right.

anyone know what to do?

$y^{\prime}(x) = abx^{b - 1}ln(x) + ax^{b - 1}$

You wish that
$y^{\prime}(e^2) = ab(e^2)^{b - 1}ln(e^2) + a(e^2)^{b - 1} = 0$

And also
$6e^{-1} = a(e^2)^b \cdot ln(e^2)$ <-- From your original equation.

These two equations simplify to
$2abe^{2(b - 1)} + ae^{2(b - 1)} = 0$
and
$\frac{6}{e} = 2ae^{2b}$

Can you solve these for a and b? (Note that you will also need to show that this is a local maximum, not a local minimum.)

-Dan

**Paige05** · November 21st 2007, 11:29 AM

those are exactly the equations that i came up with, but when i solved for a and put that back into the second equation, i find it so hard to simplify and solve for b. im stuck.

**Paige05** · November 21st 2007, 11:35 AM

nevermind! i got it!

thank you

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