Homing pigeons avoid flying over water. Suppose a homing pigeon is released on an island at point $C$, which is 11 mi directly out in the water from a point B on shore. Point $B$ is 24 mi downshore from the pigeon's home loft at point $A$. Assume that a pigeon flying over water uses energy at a rate 1.35 times the rate over land. Toward what point S downshore from A should the pigeon fly in order to minimize the total energy required to get to the home loft at A? Total energy $=($ Energy rate over water $) \cdot($ Distance over water $)+($ Energy rate over land $) \cdot($ Distance over land $)$ Point $S$ is $\square$ miles away from point $A$. (Type an integer or decimal rounded to three decimal places as needed.)

Step 1 ：The pigeon's path can be divided into two parts: the flight over water from C to some point S on the shore, and the flight over land from S to A. We can use the Pythagorean theorem to express the distance over water in terms of the distance over land. The distance over water is the hypotenuse of a right triangle with legs of lengths 11 mi (the distance from C to B) and x mi (the distance from B to S), so it is \(\sqrt{11^2 + x^2}\) mi. The distance over land is 24 - x mi (since S is x mi downshore from B and B is 24 mi downshore from A).

Step 2 ：The total energy is then \(1.35 \cdot \sqrt{11^2 + x^2} + 24 - x\). We want to minimize this expression. To do this, we can take the derivative of the expression with respect to x, set it equal to zero, and solve for x.

Step 3 ：The critical point we found, approximately 12.129, is the value of x that minimizes the total energy. This means that the pigeon should fly to a point S that is approximately 12.129 miles downshore from point A.

Step 4 ：Final Answer: Point S is \(\boxed{12.129}\) miles away from point A.