Max-Min problem

Question

A woman can run twice as fast as she can swim. She is standing at point A on the edge of a circular swimming pool 40m in diameter, and she wishes to get the diametrically opposite point B as quickly as possible. She can run around the edge to point C, the swim directly from C to B. Where should C be chosen to minimize the total time taken to get from A to B?

Answer 1

Draw a circle. Put point A on the left most side and point B on the right directly across from A. A and B are the left and right ends of the diameter.

Now put a point C at any place on the upper edge of the circle. Form an enclosed triangle A-B-C which makes

the angle at C a right angle, so A-B-C is a right triangle. Let the angle at A be θ

Draw a line from C to the center of the circle and mark it as D. By a geometry theorem, the angle at D will be twice the angle θ.

The values of the speed aren't given, but running speed is twice the swimming speed. For simplicity we can let the swimming speed be 1 m/s and the running speed be 2 m/s. You can set these to any speed, as long as

v_r=2v_s

There are three possible ways to get from A to B.

1) Swim straight across the diameter from A to B for a distance of 40 m

2) Run around the circumference of the half circle from A to B to a distance of πr = 20π m

3) Swim from A to C and run from C to B for a total distance d_s + d_r.

We want to minimize the time so we can use t=d/v, remembering to consider the boundary times.

1) t = AB/v_s = 40/1 = 40 seconds

2) t = πr/v_r = 20π/2 = 10π ≈ 31.4 seconds

3) t_total = t_s + t_r = d_s/1 + d_r/2

d_s = AC = 40cos(θ)

d_r = the length of the arc from C to B = arc length = (central angle in radians) x (radius of the circle) = 2θ(20)

= 40θ

T(θ) = t_s + t_r = d_s + d_r = 40cos(θ) + 40θ/2

T'(θ) = 40sin(θ) + 20

Setting T'(θ) to zero, sin(θ) = 1/2 --> θ = π/6 or 5π/6. 5π/6 is too big, so choose π/6

T(π/6) =40(√3/2) + 20(π/6) ≈ 34.6 + 10.4 = 45 seconds

So running around the half circle is the minimum time.

Answer 2

We can first define two variables that represent the running and swimming speeds:

Running speed: V_r [m/s]

Swimming speed: V_s [m/s]

We know that she runs twice as fast as she swims:

V_r = 2*V_s or V_s = V_r/2

We know the diameter of the pool to be 40 m. If θ (in radians) represents the angle between points A and C with respect to the center of the circle then we can calculate the arc length as follows:

Arc length (A to C) = R*θ. Here R is the radius of the pool so 20 m. 

We can then calculate the total time to run from point A to point C as follows:

T_r = (R*θ)/V_r

The distance covered by swimming from point C to point B would be the chord length. We can calculate the chord length as follows:

Chord_length = 2*R*sin(θ/2)

The total time to swim this length can be calculated as:

T_s = (2*R*sin(θ/2))/V_s

We know that V_s = V_r/2 so we can represent the time to swim equation as:

T_s = (4*R*sin(θ/2))/V_r

Now we can create an objective function that is the sum of these two times:

T_tot = T_r + T_s

Now, to minimize we need to calculate the first derivative of the objective function and find what values of θ make the derivative equal to 0. 

Once you find the optimal θ, you can use this value to determine the optimal location of C. If you were to do it in cartesian coordinates, then you could set the center of the circle at (0,0). This would mean that A is at (20,0) and B is at (-20,0). You can use the following equation to calculate the cartesian coordinates:

C = (R*cos(pi-θ), R*sin(pi-θ))

I hope this helps. Let me know if you have any further questions or would like a more detailed explanation.