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 (9,0) 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.
f(x) =
(b) We find that f(x) has a critical point at
x =
(c) To verify that f(x) has a minimum at this critical point we evaluate the second derivative f′′(x) at this point
f''(x)(critical point) is =
(d) Thus the minimum length of cable needed is
I labelled the position of Centerville, Springfield, Shelbyville and (x,0) as
C, A, B, and P respectively
Length of cable = PC + AP + BP
clearly cost is dependent on the length
Cost = PC + AP + BP , for some unit of money
= 9-x + √(x^2 + (-4)^2) + √(x^2 + 4^2)
= 9 - x + √(x^2 + 16) + √(x^2 + 16) = 2√(x^2 + 16) + 9 - x
= 2(x^2 + 16)^(1/2) + 9 - x
d(Cost)/dx = (x^2 + 16)^(-1/2) (2x) - 1
= 2x/√(x^2 + 16) - 1
= 0 for a min of Cost
2x/√(x^2 + 16) = 1
2x = √(x^2 + 16)
square both sides
4x^2 = x^2 + 16
3x^2 = 16
x = ± 4/√3 , but x must be between 0 and 9
so the point is (4/√3, 0)
I will let you decide how all that fits in to your series of questions, everything
should be there.
The only thing I didn't do was to find the 2nd derivative to establish max or min, it
was obvious.
Here is what your cost graph looks like, just need to look at 0 < x < 9 , there is a min
https://www.wolframalpha.com/input/?i=plot+y+%3D+2%E2%88%9A%28x%5E2+%2B+16%29+%2B+9+-+x
(a) To solve this problem, we need to minimize the function of x, which represents the total length of cable needed. Let's break down the length of the cable into three segments:
1. The distance from Centerville to the point (x,0) on the x-axis.
2. The distance from (x,0) to Springfield, which is the hypotenuse of a right triangle with legs of length x and 4.
3. The distance from (x,0) to Shelbyville, which is the hypotenuse of a right triangle with legs of length x and 4.
So, the total length of cable can be expressed as the sum of these three lengths:
f(x) = distance from Centerville to (x,0) + distance from (x,0) to Springfield + distance from (x,0) to Shelbyville.
(b) To find the critical point of f(x), we need to find the value of x that makes the derivative of f(x) equal to zero. Let's take the derivative of f(x):
f'(x) = derivative of distance from Centerville to (x,0) + derivative of distance from (x,0) to Springfield + derivative of distance from (x,0) to Shelbyville.
The derivative of distance from Centerville to (x,0) is 1, as it does not depend on x.
The derivative of distance from (x,0) to Springfield can be found using the Pythagorean theorem: d'(x) = sqrt(1^2 + 4^2) = sqrt(17).
The derivative of distance from (x,0) to Shelbyville is also sqrt(17).
Setting the derivative f'(x) equal to zero, we have:
1 + sqrt(17) + sqrt(17) = 0.
Simplifying, sqrt(17) = -1/2.
Since a negative square root doesn't make sense in this context, there is no solution to f'(x) = 0. Therefore, there are no critical points.
(c) Since there are no critical points, we cannot evaluate the second derivative at any critical points.
(d) Since there are no critical points, we cannot find the minimum length of cable needed. The problem does not have a solution.