(1 pt) Centerville is the headquarters of Greedy Cablevision Inc. The cable comp

Question

(1 pt) Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Springfield and Shelbyville. There needs to be cable connecting Centerville to both towns.

Centerville is located at (11,0) in the xy-plane, Springfield is at (0,2), and Shelbyville is at (0,-2). To save on the cost of cable, Greedy Cablevision wants to arrange the cable in a Y-shaped configuation, running cable from Centerville to some point (x,0) on the x-axis where it then splits into two branches, one going to Springfield and one to Shelbyville.

Find the location (x,0) that will minimize the amount of cable between the 3 towns and compute the amount of cable needed. Justify your conclusion by answering the following questions.

(a) To solve this problem we need to minimize the following function of x:

(d) Thus the minimum length of cable needed is .

Explanation / Answer

We can assume 0 <= x <= 11, because it is clear that using x > 11 or x < 0 would just waste cable.
The length of cable from Centerville at (11, 0) to (x, 0) is 11 - x, since x <= 11.
The length of cable from (x, 0) to Springfield at (0, 2) is sqrt(x^2 + 2^2).
The length of cable from (x, 0) to Shelbyville at (0, -2) is sqrt(x^2 + 2^2).

So the total length of cable is given by the function
f(x) = 11 - x + 2sqrt(x^2 + 4).

Set the derivative equal to zero:
f'(x) = -1 + 2x/sqrt(x^2 + 4) = 0.
2x/sqrt(x^2 + 4) = 1
2x = sqrt(x^2 + 4)
4x^2 = x^2 + 4
3x^2 = 4
x = sqrt(4/3), or about 1.155, since we are using x >= 0.

We now explain why the critical point x = sqrt(4/3) is a global minimum for the function f(x), on 0 <= x <= 11.

Note that since x >= 0, f'(x) = -1 + 2x/sqrt(x^2 + 4) = -1 + 2/sqrt(1 + 4/x^2) and so f'(x) always strictly increases for 0 <= x <= 11.
Since f'(sqrt(4/3)) = 0 and f'(x) always strictly increases for 0 <= x <= 11, it follows that f'(x) is negative for 0 <= x < sqrt(4/3) and is positive for sqrt(4/3) < x <= 11, as well as being zero at x = sqrt(4/3).
Therefore, the original function f(x) strictly decreases for 0 <= x < sqrt(4/3) and strictly increases for sqrt(4/3) < x <= 11, as well as having a horizontal tangent at x = sqrt(4/3).
So x = sqrt(4/3) is a global minimum for the function f(x), on 0 <= x <= 11.

The minimum length of cable is then
f(sqrt(4/3)) = 11 - sqrt(4/3) +2sqrt(4/3 + 4) = 14.464 .

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