Optimization problem, not sure how to proceed

Question

So I'm a bit confused by this optimization word problem. I would be able to solve it I think given number values for the speeds but I'm uncertain how to get an exact answer when you don't know the difference between $v_1$ and $v_2$. Maybe I'm just not sure how to approach the question.

You stand adjacent to point $A$ at the edge of a circular lake of radius $r$. Want to get to $B$ diametrically opposite $A$. You can swim straight, run around the shore, or swim to a point $C$ somewhere in between and run the rest of the way to $B$.

Assume running speed $v_1$, swimming speed $v_2$, with $v_2 < v_1$.

Answer 1

Since the lake is circular, we can use angles to determine the position of the person and the points. Let point A be at $0 ^{\circ}$ and point B be at $180 ^{\circ}$

Note that the segment between point A and point C in the circle is a chord, and its length can easily be found and proven as well.

I would approach this problem by creating a time function based on the angle between point A, center of the lake, and some point C at the edge of a lake. This function measures the total time for the person to reach from point A to the midpoint of the lake.

There would be two components to this function, the time it took for the person to swim and the time it took for the person to run. These two components are also based on the angle of the function. There is a relationship to these two components based on the angle: the longer the person swims, the shorter amount s/he has to run around the lake to the midpoint.

After you generate a time function that reflects the two components and how they're related, it's simply taking a matter of the function's derivative to find the absolute minima of the time.

The time it takes for the person to swim and get to shore at some point C would be something like $(r*2*sin(\theta/2))/v_2$. The numerator is a well known length of the chord between point A and C in the circle. Notice that if the person decided not to swim at all, then point C = point A and $\theta = 0$ and this component $= 0$ as well.

The time it takes for the person to then run from point C to point B would then be: $2*r *(\pi - \theta)/v_1$. Notice that if the person decided not to run at all, then point C = point B and $\theta = \pi$ and this component $=0$ as well.

Therefore, the function that determines the time for the person to go from point A to point B, with the possibility of point C in between, would be something like $t(\theta) = (2r*sin(\theta/2))/v_2 + (2r * (\pi - \theta)/v_1)$

I say "something like" because I didn't critique this equation much, but the idea behind this is correct.

Answer 2

This is an example of what I regard as a "sneaky" optimization problem, because it turns out that the optimal choice is not given by the "critical point" in the interval.

We have the choices of: (1) swimming along the diameter $ \ AB \ $ ; (2) running along half the circumference of the circle from $ \ A \ $ to $ \ B \ $ ; or (3) choosing a point $ \ C \ $ on the shore to which to swim from $ \ A \ $ , and then to run along the shore from $ \ C \ $ to $ \ B \ $ .

BirdKiller1989 describes a good approach to setting up the analysis for finding the time required, starting from a point $ \ A \ $ to reach the "far side" of the circular lake of radius $ \ R \ $ at point $ \ B \ $ . A convenient variable to use in our work on is the "central angle" $ \ \angle \ AOC \ $ , which has the measure $ \ \theta \ $ in the interval $ \ 0 \ \le \ \theta \ \le \ \pi \ $ . It will be useful to know the length of the chord $ \ AC \ $ of the circle, which we can calculate from, for instance, the Law of Cosines (or we could just look it up), as

$$ \ s \ = \ \sqrt{2R^2 \ - \ 2R^2 \ \cos \ \theta} \ = \ \sqrt{2} \ R \ \sqrt{1 \ - \ \cos \ \theta} \ \ . $$

We shall also want the measure of the other central angle, $ \ \angle \ COB \ $ , which is, in radians, $ \ \pi \ - \ \theta \ $ . The distance run along the shore from $ \ C \ $ to $ \ B \ $ is then $ \ d \ = \ R \ ( \ \pi \ - \ \theta \ ) \ $ .

We have two unspecified quantities still to deal with, which are the swimming speed $ \ v_2 \ $ and the running speed $ \ v_1 \ $ . It will make comparisons easier if we write $ \ v_1 \ = \ k \cdot v_2 \ , \ k \ > \ 1 \ $ .

Choice (1) then requires a time of $ \ T_1 \ = \ \frac{2R}{v_2} \ $ , while choice (2) requires $ \ T_2 \ = \ \frac{\pi R}{v_1} \ = \ \frac{\pi R}{k v_2} \ $ . We shall want to keep these results in mind.

Choice (3), the "mixed" solution of swimming and running, will need a total time of

$$ \ T_3 \ = \ \frac{s}{v_2} \ + \ \frac{d}{v_1} \ = \ \frac{\sqrt{2} \ R \ \sqrt{1 \ - \ \cos \ \theta}}{v_2} \ + \ \frac{R \ ( \ \pi \ - \ \theta \ )}{kv_2} $$

$$ = \ \frac{R}{v_2} \left[ \ \sqrt{2} \ (1 \ - \ \cos \ \theta)^{1/2} \ + \ \frac{( \ \pi \ - \ \theta \ )}{k} \ \right] \ \ . $$

It is this function that we shall wish to examine for critical points on the interval $ \ 0 \ < \ \theta \ < \ \pi \ $ , with the possibility that the result will depend in some important way on the value of $ \ k \ $ . Upon differentiating this function with respect to $ \ \theta \ $ , we obtain

$$ \ \frac{d \ T_3}{d\theta} \ = \ \frac{R}{v_2} \left[ \ \sqrt{2} \ \cdot \ \frac{1}{2} \ (1 \ - \ \cos \ \theta)^{-1/2} \ \cdot \ \sin \ \theta \ - \ \frac{1}{k} \ \right] \ \ ; $$

we can make this a little easier to work with by applying a "conjugate factor" to write

$$ \frac{\sin \ \theta}{\sqrt{1 \ - \ \cos \ \theta}} \ \cdot \ \frac{\sqrt{1 \ + \ \cos \ \theta}}{\sqrt{1 \ + \ \cos \ \theta}} \ = \ \frac{\sin \ \theta \ \sqrt{1 \ + \ \cos \ \theta}}{\sqrt{1 \ - \ \cos^2 \theta}} \ = \ \sqrt{1 \ + \ \cos \ \theta} \ \ , $$

(which is equal to the original ratio on the interval $ \ 0 \ < \ \theta \ < \ \pi \ $ ) , leading to

$$ \ \frac{d \ T_3}{d\theta} \ = \ \frac{R}{v_2} \left[ \ \frac{\sqrt{2}}{2} \ (1 \ + \ \cos \ \theta)^{1/2} \ - \ \frac{1}{k} \ \right] \ \ . $$

Now, it would seem, we want to set this derivative equal to zero and solve for the optimal value of $ \ \theta \ $ . However, we won't bother doing that: as we discover by taking the next derivative (or by graphing $ \ T_3 \ $ for various values of $ \ k \ $ ) ,

$$ \ \frac{d^2 T_3}{d\theta^2} \ = \ \frac{R}{v_2} \ \cdot \ \frac{\sqrt{2}}{2} \ (1 \ + \ \cos \ \theta)^{-1/2} \ \cdot \ ( - \sin \ \theta ) \ = \ -\frac{R \ \sqrt{2}}{2v_2} \ \frac{\sin \ \theta}{\sqrt{1 \ + \ \cos \ \theta}} \ \ . $$

Since both the numerator and the denominator in the trigonometric ratio are positive on the interval $ \ 0 \ < \ \theta \ < \ \pi \ $ , the second derivative is negative everywhere on the interval, meaning that the critical point for the function $ \ T_3 (\theta) \ $ is a maximum. Moreover, there is no dependence of the second derivative upon $ \ k \ $ ; this tells us then that choice (3) is never the proper choice for passing from $ \ A \ $ to $ \ B \ $ in the shortest possible time.

This is a situation where what we learn in first-semester calculus as the "Extreme Value Theorem" becomes useful: a continuous function on a closed interval has an absolute maximum value and an absolute minimum value. Since the critical point in the open interval is not minimal, that absolute minimum must lie at one of the endpoints, either $ \ \theta \ = \ \pi \ $ (all swimming) or $ \ \theta \ = \ 0 \ $ (all running).

So here is where we recall our times for choices (1) and (2) ; by comparing them, we have

$$ \frac{T_1}{T_2} \ = \ \frac{2R \ / \ v_2}{\pi R \ / \ k v_2} \ = \ \frac{2k}{\pi} \ \ . $$

It takes equally long to run or swim all the way ( $ \ \frac{T_1}{T_2} \ = \ 1 \ $ ) for $ \ k \ = \ \frac{\pi}{2} \ \approx \ 1.57 $ ; if $ \ k \ < \ \frac{\pi}{2} \ $ , it will take less time to swim the diameter of the lake; otherwise, it's quicker to run along the semi-circle. Since the "typical person" has a (sustained) running speed-to-swimming speed ratio of $ \ k \ > \ 4 \ $ , it would take less time for most people to run around the lake.