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. The idea is to save on the cost of cable by arranging the cable in a Y-shaped configuation. Centerville is located at (8,0) in the xy-plane, Springfield is at (0,7), and Shelbyville is at (0,- 7). The cable runs from Centerville to some point (x,0) on the x-axis where it splits into two branches going to Springfield and 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 answer.
To solve this problem we need to minimize the following function of x:
f(x)=______________
We find that f(x) has a minimum at x=_____________
Thus the minimum length of cable needed is _______________
To solve this problem we need to minimize the following function of x:
f(x)=______________
We find that f(x) has a minimum at x=_____________
Thus the minimum length of cable needed is _______________
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We can assume 0 <= x <= 8, because it is clear that using x > 8 or x < 0 would just waste cable.
The length of cable from Centerville at (8, 0) to (x, 0) is 8 - x, since x <= 8.
The length of cable from (x, 0) to Springfield at (0, 7) is sqrt(x^2 + 7^2).
The length of cable from (x, 0) to Shelbyville at (0, -7) is sqrt(x^2 + 7^2).
So the total length of cable is given by the function
f(x) = 8 - x + 2sqrt(x^2 + 49).
Set the derivative equal to zero:
f'(x) = -1 + 2x/sqrt(x^2 + 49) = 0.
2x/sqrt(x^2 + 49) = 1
2x = sqrt(x^2 + 49)
4x^2 = x^2 + 49
3x^2 = 49
x = sqrt(49/3), or about 4.041, since we are using x >= 0.
We now explain why the critical point x = sqrt(49/3) is a global minimum for the function f(x), on 0 <= x <= 8.
Note that since x >= 0, f'(x) = -1 + 2x/sqrt(x^2 + 49) = -1 + 2/sqrt(1 + 49/x^2) and so f'(x) always strictly increases for 0 <= x <= 8.
Since f'(sqrt(49/3)) = 0 and f'(x) always strictly increases for 0 <= x <= 8, it follows that f'(x) is negative for 0 <= x < sqrt(49/3) and is positive for sqrt(49/3) < x <= 8, as well as being zero at x = sqrt(49/3).
Therefore, the original function f(x) strictly decreases for 0 <= x < sqrt(49/3) and strictly increases for sqrt(49/3) < x <= 8, as well as having a horizontal tangent at x = sqrt(49/3).
So x = sqrt(49/3) is a global minimum for the function f(x), on 0 <= x <= 8.
The minimum length of cable is then
f(sqrt(49/3)) = 8 - sqrt(49/3) +2sqrt(49/3 + 49) = 8 + 7sqrt(3) or about 20.124 .
Lord bless you today!
The length of cable from Centerville at (8, 0) to (x, 0) is 8 - x, since x <= 8.
The length of cable from (x, 0) to Springfield at (0, 7) is sqrt(x^2 + 7^2).
The length of cable from (x, 0) to Shelbyville at (0, -7) is sqrt(x^2 + 7^2).
So the total length of cable is given by the function
f(x) = 8 - x + 2sqrt(x^2 + 49).
Set the derivative equal to zero:
f'(x) = -1 + 2x/sqrt(x^2 + 49) = 0.
2x/sqrt(x^2 + 49) = 1
2x = sqrt(x^2 + 49)
4x^2 = x^2 + 49
3x^2 = 49
x = sqrt(49/3), or about 4.041, since we are using x >= 0.
We now explain why the critical point x = sqrt(49/3) is a global minimum for the function f(x), on 0 <= x <= 8.
Note that since x >= 0, f'(x) = -1 + 2x/sqrt(x^2 + 49) = -1 + 2/sqrt(1 + 49/x^2) and so f'(x) always strictly increases for 0 <= x <= 8.
Since f'(sqrt(49/3)) = 0 and f'(x) always strictly increases for 0 <= x <= 8, it follows that f'(x) is negative for 0 <= x < sqrt(49/3) and is positive for sqrt(49/3) < x <= 8, as well as being zero at x = sqrt(49/3).
Therefore, the original function f(x) strictly decreases for 0 <= x < sqrt(49/3) and strictly increases for sqrt(49/3) < x <= 8, as well as having a horizontal tangent at x = sqrt(49/3).
So x = sqrt(49/3) is a global minimum for the function f(x), on 0 <= x <= 8.
The minimum length of cable is then
f(sqrt(49/3)) = 8 - sqrt(49/3) +2sqrt(49/3 + 49) = 8 + 7sqrt(3) or about 20.124 .
Lord bless you today!