Optimization problem: rowing across a lake 
A woman at a point A on the shore of a circular lake with radius $r=3$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time. She can walk at the rate of 10 mph and row a boat at 5 mph. What is the shortest amount of time it would take her to reach point $C$?


My Working
Let $C\hat{A}B=\theta$
$\begin{align*}\therefore AB+arcBC&=\sqrt{3^2+3^2-2\cdot3\cdot3\cdot\cos(\pi - 2\theta)}+3\cdot2\theta\\
&=3\sqrt{2-2\cos(\pi-2\theta)}+6\theta\end{align*}$
$\begin{align*}\therefore \text{time taken} =t=\frac{3\sqrt{2-2\cos(\pi-2\theta)}}{5}+\frac{6\theta}{10}\end{align*}$
To find the point $B$ at which $t$ is minimized, we make $\frac{\operatorname d t}{\operatorname d\theta}=0$:
$\begin{align*}\frac{\operatorname d t}{\operatorname d\theta}=\frac{3-4\sin(\pi-2\theta)}{10\sqrt{2-2\cos(\pi-2\theta)}}+\frac{3}{5}=0\end{align*}$
And the rest of my working gets messy from here. Is there a better method than this?
 A: If you call the center of the circle $O$ and let $\alpha=\angle AOB$, then the woman's rowing distance is $2r\sin \frac{\alpha}{2}=6\sin \frac{\alpha}{2}$; and her walking distance is $r\cdot (\pi-\alpha)=3(\pi-\alpha)$.
So her travelling time is $T(\alpha)=\frac{6}{5}\sin\frac{\alpha}{2}+\frac{3}{10}(\pi-\alpha)$.
I think you will find that this minimizes nicely with calculus methods.
A: I would like to propose a calculus approach here. So the question is vague in the sense that, we don't really know if she has to walk first or row first. But it's actually symmetric in nature, if you consider walking (along the arc) first (say 'Q' miles) AND then row the remaining distance (say 'P' miles), it's literally symmetric to rowing the same previous distance ('P' miles) first followed by walking ('Q' miles).
Consider any case among these as it doesn't matter due to the above proof. I'll take latter one here. So, assuming $\angle{BAC}=\theta $, we can conclude few things, the 'P' meters becomes $ 2r \times cos(\theta)$ miles, and the 'Q' miles of arc length corresponds to $r \times 2 \theta$ miles. So, now plugging in the value for r as 3, we have the total time as $T_{total}= (\frac{2 \times \cos(\theta)}{5} + \frac{3 \times 2 \times \theta}{10}) \text{hrs}$.
Now we just have to find where this function has the least value. But we have to be sure of the domain we consider here, as we HAVE to reach the other side, we could start with $\theta=0$ or have a maximum angle of $\frac{\pi}{2}$ rad. (We cannot have $\theta > \pi$ because that would mean we're not wanting to reach the other side.)
Keeping these things in mind, we find that the critical point occurs at $\theta$ = 0.5235. Now this value if substituted in the total time equation, gives us the time as $\approx$ 1.667 hr. Now let's check how the function behaves at the end points in the domain (remember, we can include 0 and $\frac{\pi}{2}$ rads). $\text{T}_{\theta=0}= 1.2 hr$ and $\text{T}_{\theta=\frac{\pi}{2}} \approx 0.942$. So, turns out that if you just first row for 0 miles at an angle of $90^{o}$ and then walk for the remaining $(3 \times 2 \pi)$ miles, it takes the least amount of time.
