Answer:
20.2057 Units.
Step-by-step explanation:
First, we determine the length of the cable.
Distance between Centerville (8,0) and point (x,0) is given as:
Distance between point (x,0) and Springfield(0,7) is:
Distance between point (x,0) and Shelvyfield(0,-7) is:
Therefore the Length of the Cable L(x)
To find the critical point, we set the derivative of L(x)=0


To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

Since
is positive, the minimum point of L(x) exists.
Next, we find the minimum length by substituting z=3.1305 into L(x)

Minimum Length, L=20.2057
The minimum length of the cable is 20.2057 Units.