Ecological Ed must cross a circular lake of radius 1 mile... This is a basic single variable optimisation problem from chapter 11 of Spivak's Calculus:


*Ecological Ed must cross a circular lake of radius 1 mile. He can row across at 2 mph or walk around at 4 mph, or he can row part way and walk the rest (Figure 28). What route should he take so as to;

a) See as much scenery as possible?
b) Cross as quickly as possible?

First of all, I really don't get part a. Am I supposed to maximise the distance of Ed's route? If so, why would I need calculus for that, obviously the maximum distance would be walking the whole way round the lake?
Part b is a little more interesting, but the solution I got was that Ed should walk the whole distance, which makes me think I may have made a mistake somewhere as usually in these types of textbook problems the solution is between the possible extremities, not at one of them (Although I suppose that's a good reason to include such a question, to throw off the student a little!).
My setup was to represent the circle as the graph of $x^2 + y^2 = 1$, then let $C$ be the point at which Ed switches from rowing to walking, somewhere along the circumference. Therefore $C = (\cos(c),\sin(c))$ where $c \in [0, \pi]$ represents the angle between $OC$ and the x-axis (The same way sin and cos are defined in the first place). For example; when $c = 0$, then $C = (1,0)$, which would correspond to when Ed rows the whole way to $(1,0)$. When $c = \pi$, then $C = (-1,0)$, which corresponds to when Ed rows nowhere and walks the (entire) rest of the way. All  other intermediate values of $c$ will give when Ed rows a bit, then walks the rest, as the question asks.
Anyway, skipping the details now, I got the function for the total time Ed's route takes given the variable $c$ is $t(c) = \frac{1}{4}(c+2\sqrt{2}\sqrt{\cos(c)+1})$. Is this correct? I'm pretty sure it is. But I'm even more sure that if it is correct, then the minimum of $t(c)$ occurs when $c= \pi$, which means Ed should walk the whole way? Can someone confirm/de-confirm?
 A: I'd interpret "seeing as much scenery" to mean the longest distance. This is easily seen to be walking around the lake.
Your solution for part b is correct. Another way to see this is to directly compare rowing across part of the lake with walking that same arclength around. Say the angle of the lake covered from the starting point is $\theta$. Then the walking distance is $\underbrace{\frac{\theta}{2\pi}}_{\text{fraction of lake}} \cdot \underbrace{2\pi \text{ miles}}_{\text{circumference of lake}} = \theta \text{ miles}$. The time for this would be $\frac{\theta \text{ miles}}{4 \text{ mph}} = \frac{\theta}{4} \text{ hours}$.
Alternately, you could directly row across to that point. The distance (using the law of cosines) would be $\sqrt{2-2\cos(\theta)}$ miles. The time for this would be $\frac{\sqrt{2-2\cos(\theta)} \text{ miles}}{2 \text{ mph}} = \frac{\sqrt{2-2\cos(\theta)}}{2} \text{ hours} = \sin\left( \frac{\theta}{2} \right) \text{ hours}$. For $\theta \in [0, \pi]$, this would always be greater than the walking time.
A: a. to see maximum scenery, Ed can travel a full turn, then cross by any route ! (No other optimization criterion is requested.)
b. if I am right, the tour must be made of a single, straight row segment. Let the central angle subtended by this segment be $2\alpha$. The total time is
$$\frac{\pi-2\alpha}4+\frac{2\sin\alpha}2.$$
Taking the derivative, the minimum is achieved when
$$\cos\alpha=\frac12.$$
Hence the optimal path is made of a side of an inscribed hexagon and one or two arcs.
