Prove that is the minimum of a function An object with weight $W$ is dragged along a horizontal plane by a force acting a rope attached to the objetc. If the rope makes an angle $\theta$ with the plane then magnitude of the force is $$F = \dfrac{\mu W}{\mu \sin \theta + \cos \theta}$$ where $\mu$ is a positive constant called coefficient friction and where $0 \leq \theta \leq \dfrac{\pi}{2}$. Show that $F$ is minimized when $\tan \theta = \mu$.
What i tried: $$F'(\theta) = \dfrac{-\mu W (\mu \cos \theta - \sin \theta)}{(\mu \sin \theta + \cos \theta)^{2}}$$ $$F'(\theta) = 0 \Leftrightarrow -\mu W (\mu \cos \theta - \sin \theta) = 0$$ $$\mu^{2}W\cos \theta = \mu W \sin \theta$$ As $\mu > 0$ and considering $\theta \neq \dfrac{\pi}{2}$ we can write : $$\mu = \tan \theta (*)$$
I know that: 
1) $F(0) = \mu W$
2) $F(\dfrac{\pi}{2}) = W$
From (*), we have : $\theta = \arctan \mu$
How can I conclude that $F(\arctan \mu) < F(0)$ and $F(\arctan \mu) < F(\dfrac{\pi}{2})$?
thanks in advance.
 A: With $\tan \theta = \mu$, you get,
$$\cos\theta = \frac {1}{\sec\theta} = \frac{1}{\sqrt{1+\tan^2\theta}} = \frac{1}{\sqrt{1+\mu^2}}$$
$$\sin\theta = \tan\theta\cos\theta = \frac{\mu}{\sqrt{1+\mu^2}}$$
Plug them into 
$$F = \dfrac{\mu W}{\mu \sin \theta + \cos \theta}$$
to get
$$F(\arctan\mu) = \frac{\mu W}{\sqrt{1+\mu^2}}$$
Since $\frac{\mu }{\sqrt{1+\mu^2}}<1$, you can conclude,
$$ F(\arctan \mu) = \frac{\mu }{\sqrt{1+\mu^2}}W < W = F\left(\frac{\pi}{2} \right) $$
Also, since $\frac{1}{\sqrt{1+\mu^2}}< 1$, you can conclude,
$$ F(\arctan \mu) = \frac{1}{\sqrt{1+\mu^2}} \mu W< \mu W = F(0) $$
A: Here's a different approach.
The denominator of $F$ is smooth and positive (as you can check) and the numerator is a positive constant.
Hence $F>0$ and you can just maximize the denominator $D$ to minimize the whole fraction. Computing $D'=0$ you'll get $\mu=\tan(\theta)$.
Now you can check that $D''=-D<0$ so that you have a local maximum of the denominator.  You should also check that both $D(\pi/2)$ and $D(0)$ are strictly smaller than $D(\arctan(\mu))$, then you'll get you answer.
