Question

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 configuration.

Centerville is located at (12,0) in the x -plane, Springfield is at (0,6) , and Shelbyville is at (0,−6) . 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 critical number at x = ?
To verify that f(x) has a minimum at this critical number we compute the second derivative f''(x) and find that its value at the critical number is ? , a positive number.
Thus the minimum length of cable needed is ?

Answer 1

x = 3,46

d = 22,38 m

Step-by-step explanation: See Annex

According to the problem description.

Let´s call L₁ total distance between Centerville and Springfield

and L₂ total distance between Centerville and Shelbyville

Then, the total cable will be:

L₁ + L₂ note that the length 12 - x ( in red in the annex ) is common to the two distances (we only need to take it once)

Therefore total distance d = ( 12 - x ) + 2 √(x² ) + (6)²

F(x) = 12 - x + 2*√ ( x² + 36 )

Taking derivatives on both sides of the equation

F´(x) = -1 + 2 * 1/2 * 2*x/ √( x² + 36 )

F´(x) = - 1 + 2*x / √(x² + 36 )

F´(x) = 0 -1 + 2*x/ √( x² + 36 ) = 0

Solving for x

- √ ( x² + 36 ) + 2*x = 0

- √ ( x² + 36 ) = - 2*x

√ ( x² + 36 ) = 2*x

squaring both sides

x² + 36 = 4 x²

3*x² = 36

x² = 12

x = 3,46 m then ( 3,46 , 0 ) is the location for minimun length

Then total amount of cable is:

d = 12 - 3,46 + 2 * √ (3,46)² + 36

d = 8,54 + 2 * 6,92

d = 22,38 m

if we take the second derivative of F

F´´(x) = 0 + D/ dx ( 2*x/ √( x² + 36 ) )

F´´(x) = [ 2*√( x² + 36 ) - 2*x [ x/√ ( x² + 36 ) ] / (x² + 36)

we can see that this expression is an integer positive, since the second term is always smaller than the first one

then we have a minimun for x = 3,46