The distance that you will swim will be the third leg of an isosceles triangle. Splitting this triangle down the middle gives two congruent triangles with hypotenuse of 1 mile. Let's call the base of each of these triangles 'x'. When you draw it, you'll notice that:
x/1 = cosΘ ==> 2x = 2cosΘ.
Therefore, the distance that you will swim is 2cosΘ, since there are two triangles. The rate that you'll travel this distance is 'r'.
This distance that you'll jog is pi minus the arc-length subtended by the angle pi - 2Θ.
(pi - (pi - 2Θ)) = 2Θ
The rate that you'll jog this distance is 2r.
Now let's use: time = distance / rate.
t = 2cosΘ/r + 2Θ/2r.
Differentiating yields:
t ' = -2sinΘ/r + 1/r.
Now let's set our numerator equal to 0.
-2sinΘ + 1 = 0
sinΘ = 1/2
Θ = pi/6
t(pi/6) = 2cos(pi/6)/r + 2(pi/6)/2r
t(pi/6) = (sqrt3)/r + pi/6r.
Since you didn't give the rate, I can't find the exact time in minutes. But notice that the rate at which he swims must be slow. Let's say that it is 1/10 mi/min. Then we'd have:
t(pi/6) = (sqrt3)/(1/10) + pi(6/10)
t(pi/6) = 17.3 + 5.23 = 22.5.
Therefore, the time is reasonable.
Essentially, you'll swim a distance of sqrt3 miles to the point on the circle that makes an angle of 2pi/3 with the point that you started from. Then you'll jog a distance of pi/3 around the edge of the lake. Note that the straight-line distance is 2 miles, since radius is 1. The distance that you'll travel, however is pi/3 + sqrt3 = 2.77 miles, which makes sense.
Hope this helped.