Optimization problem -- Trouble differentiating function

[Eonfluxx]

1. The problem statement, all variables and given/known data
The efficiency of a screw, E, is given by
E=\frac{(\Theta - \mu\Theta^{2})}{\mu + \Theta} , \Theta > 0
where \Theta is the angle of pitch of the thread and \mu is the coefficient of friction of the material, a (positive) constant. What value of \Theta maximizes E?


2. Relevant equations
\frac{f'g-g'f}{g^{2}}


3. The attempt at a solution
I know what to do once I have the derivative of the function, but for this particular function I've had trouble using the quotient rule (I didn't know how to turn it into the product rule, that would have been nice too). I came up with two possibilities and any help would be appreciated:
\frac{dE}{d\Theta} = \frac{\mu - \mu^{2}2\Theta - \mu2\Theta^{2} - \Theta + \mu\Theta^{2}}{(\mu + \Theta)^{2}}

and I also came up with

\frac{dE}{d\Theta}= \frac{-\mu2\Theta - 2\Theta^{2}}{(\mu + \Theta)^{2}}

any help on which could be correct (or both wrong) would be greatly appreciated. I've read these forums a lot before and finally have a question of my own. Thanks for reading/help.

[Mark44]

I don't get either one.
\frac{d}{d\theta}\left(\frac{\theta - \mu \theta ^2}{\mu + \theta}\right)~=~\frac{(\mu + \theta)(1 - 2\mu \theta) - (\theta - \mu \theta ^2)(1)}{(\mu + \theta)^2}~=~\frac{\mu - 2\mu^2 \theta - \mu \theta^2}{(\mu + \theta)^2}

Man, that's a lot of work to get all those Greek letters looking good!

[Eonfluxx]

Bah, I made a distributing error which would result in a -\Theta and +\Theta canceling out, thanks a lot!
And thanks, it makes it easier to read

Now when setting \frac{dE}{d\Theta} equal to zero I came up with:

\Theta = \frac{-\mu}{-2\mu^{2} - \mu\Theta}

Is this correct? Thanks again for reading/help

[Mark44]

No, for the reason that you haven't really solved for theta - it's still present on the right side of your equation.
\mu - 2\mu^2 \theta - \mu \theta^2~=~ 0
\Rightarrow \mu (1 - 2\mu \theta - \theta^2)~=~ 0

So either \mu = 0 or the quadratic is 0. Use the quadratic formula to solve
\theta^2 + 2\mu \theta - 1 ~=~ 0
(I switched the signs to make solving it easier. Both equations have the same solutions.)

[Eonfluxx]

Got it, thanks again! I'll be back in the future