Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 64

$\displaystyle F = \frac{\mu W}{\mu \sin \theta + \cos \theta}; \quad 0 \leq \theta \leq \frac{\pi}{2}$ where $\mu$ is a positive constant called the coefficient of friction.
$W$ represents the weight of an object.
$\theta =$ angle
$F =$ force
Show that $F$ is minmized when $\tan \theta = \mu$.

Taking the derivative of the function,
$\displaystyle \frac{dF}{d\theta} = \mu W \cdot \frac{d}{d\theta} \left( \frac{1}{\mu \sin \theta + \cos \theta} \right)$

Using Quotient Rule,


$\begin{equation} \begin{aligned} \frac{dF}{d \theta} &= \mu W \left[ \frac{\mu\sin\theta + \cos \theta \cdot \frac{d}{d\theta} (1) - 1 \cdot \frac{d}{d\theta}(\mu \sin \theta + \cos \theta) }{(\mu \sin \theta + \cos \theta)^2}\right]\\ \\ \frac{dF}{d \theta} &= \mu W \left[ \frac{0 - \left(\mu \cos \theta + (-\sin \theta) \right)}{(\mu \sin \theta + \cos \theta)^2} \right]\\ \\ \frac{dF}{d \theta} &= \frac{\mu W \left( \mu \cos \theta + (-\sin \theta) \right)}{(\mu \sin \theta + \cos \theta)^2} \end{aligned} \end{equation}$


When $\displaystyle \frac{dF}{d\theta} = 0$,


$\begin{equation} \begin{aligned} 0 & = \frac{-\mu W (\mu \cos \theta - \sin \theta) }{(\mu \sin \theta + \cos \theta)^2}\\ \\ 0 &= \mu \cos \theta - \sin \theta\\ \\ \frac{\sin \theta}{\cos \theta} &= \frac{\mu \cancel{\cos \theta}}{\cancel{\cos \theta}} \\ \tan \theta &= \mu \end{aligned} \end{equation}$


It shows that $F$ is minimized when $\tan \theta = \mu$