## Calculus: Early Transcendentals 8th Edition

$F$ is the smallest when $\theta = tan^{-1}~\mu$
$F = \frac{\mu~W}{\mu~sin~\theta+cos~\theta}$ Note that $0 \leq \theta \leq \frac{\pi}{2}$ We can find the value of $\theta$ when $F'(\theta) = 0$: $F(\theta) = \frac{\mu~W}{\mu~sin~\theta+cos~\theta}$ $F'(\theta) = \frac{-(\mu~W)(\mu~cos~\theta-sin~\theta)}{(\mu~sin~\theta+cos~\theta)^2} = 0$ $\mu~W~sin~\theta = \mu^2~W~cos~\theta$ $sin~\theta = \mu~cos~\theta$ $tan~\theta = \mu$ $\theta = tan^{-1}~\mu$ When $tan~\theta = \mu$: $sin~\theta = \frac{\mu}{\sqrt{\mu^2+1}}$ $cos~\theta = \frac{1}{\sqrt{\mu^2+1}}$ We can find the value of $F$ when $tan~\theta = \mu$: $F =\frac{\mu~W}{\mu~sin~\theta+cos~\theta}$ $F =\frac{\mu~W}{\mu\cdot\frac{\mu}{\sqrt{\mu^2+1}}+\frac{1}{\sqrt{\mu^2+1}}}$ $F =\frac{\mu~W~\sqrt{\mu^2+1}}{\mu^2+1}$ $F =\frac{\mu~W}{\sqrt{\mu^2+1}}$ We can verify the value of $F$ at the endpoints of the interval $[0, \frac{\pi}{2}]$: $F(0) =\frac{\mu~W}{\mu~sin~0+cos~0} = \mu~W$ $F(\frac{\pi}{2}) =\frac{\mu~W}{\mu~sin~(\pi/2)+cos~(\pi/2)} = W$ Note that when $tan~\theta = \mu$: $F =\frac{\mu~W}{\sqrt{\mu^2+1}} \lt \frac{\mu~W}{\sqrt{1}} = \mu~W$ $F =\frac{\mu~W}{\sqrt{\mu^2+1}} \lt \frac{\mu~W}{\sqrt{\mu^2}} = W$ Therefore, $F$ is the smallest when $\theta = tan^{-1}~\mu$