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 (10,0) in the xy-plane, Springfield is at (0,4), and Shelbyville is at (0,-4). 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:

b. We find that f(x) has a critical point at:

c. To verify that f(x) has a minimum at this critical point we evaluate the second derivative f''(x) at this point.f''(critical point) is:

d. Thus the minimum length of cable needed is:

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  1. As usual, draw a diagram, and it is clear that

    f(x) = (10-x) + 2√(x^2+16)
    f'(x) = -1 + 2x/√(x^2+16)
    f"(x) = 32/(x^2+16)^3/2

    Now the answers are easy to provide

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  2. Thank you! I was able to find everything other than the minimum length of the cable. How would you find that?

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  3. Am I correct on these?

    1.Does every line have both an x-intercept and a y-intercept?

    2.Does every line have an infinite number of lines that are parallel to the given line?

    Does every line have an infinite number of lines that are perpendicular to the given line?

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  4. the minimum cable length occurs where f'(x) = 0

    So, you need

    2x/√(x^2+16) - 1 = 0
    2x - √(x^2+16) = 0
    2x = √(x^2+16)
    4x^2 = x^2+16
    3x^2 = 16
    x = 4/√3

    So, the minimum cable length is

    f(4/√3) = 10+4√3

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  5. vertical lines have no y-intercept
    horizontal lines have no x-intercept
    (unless these lines are in fact x=0 or y=0)

    #2 ok

    #3 also yes. Pick any point on the line (and there are an infinite number of points) and you can draw a perpendicular line through it.

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  6. this problem makes me want to di

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