Romeo and Juliet

Romeo's Path

Take a look the picture above. Let point A be the Romeo's house, P be the point where the boat reaches the shore, and point B be the Juliet's house. Since Romeo must row and walk in order to reach Juliet's house, the path that he will take is rowing a boat from point A to point P and walking from point P to point B . the time taken from point A to point B can be expressed as a function of $\theta$ for $0\le\theta\le\frac{\pi}{2}$ . $$ $\begin{aligned} T&=t_{AP}+t_{PB}\\ &=\frac{x}{3}+\frac{s}{6}\\ &=\frac{2\cos\theta}{3}+\frac{2\theta}{6}\\ &=\frac{2}{3}\cos\theta+\frac{\theta}{3}. \end{aligned}$ $$ To obtain the maximum value of $T$ , we take first derivative of $T$ with respect to $\theta$ and set equal to zero. We have $$ $\begin{aligned} \frac{dT}{d\theta}&=\frac{d}{d\theta}\left(\frac{2}{3}\cos\theta+\frac{\theta}{3}\right)\\ 0&=-\frac{2}{3}\sin\theta+\frac{1}{3}\\ \sin\theta&=\frac{1}{3}\\ \theta&=\frac{\pi}{6}. \end{aligned}$ $$ Thus, the longest possible time in minute for Romeo to get Juliet's house is $$ $\begin{aligned} T&=\frac{2}{3}\cos\left(\frac{\pi}{6}\right)+\frac{1}{3}\cdot\frac{\pi}{6}\\ &=\frac{2}{3}\cdot\frac{\sqrt{3}}{2}+\frac{\pi}{18}\\ &=\left(\frac{6\sqrt{3}+\pi}{18}\right)\text{ h}\\ &=\left(\frac{6\sqrt{3}+\pi}{18}\right)\cdot60\text{ min}\\ &\approx45.112992\\ \lfloor \text{T}\rceil&=\boxed{45\text{ min}} \end{aligned}$