Let the location of the hub(the split point) be (x,0).

Therefore the distances from the each town can eb calculated as follows,

Centerville, (9,0)

`d1 = 9-x`

Springfield (0,3)

`d2 = sqrt(3^2+x^2)`

`d2 = sqrt(9+x^2)`

Shelbyville (0,-3) same as above,

`d3 = sqrt(9+x^2)`

A)The total length of cable f(x) is...

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Let the location of the hub(the split point) be (x,0).

Therefore the distances from the each town can eb calculated as follows,

Centerville, (9,0)

`d1 = 9-x`

Springfield (0,3)

`d2 = sqrt(3^2+x^2)`

`d2 = sqrt(9+x^2)`

Shelbyville (0,-3) same as above,

`d3 = sqrt(9+x^2)`

A)The total length of cable f(x) is d1+d2+d3

`f(x) = (9-x)+sqrt(9+x^2)+sqrt(9+x^2)`

`f(x) = (9-x)+2sqrt(9+x^2)`

B) To check for the critical point, we need to calculate the first derivative of f(x)

`(df(x))/(dx) = 0 + (-1) + 2*(1/2)*2*x*(1/sqrt(9+x^2))`

`(df(x))/(dx) = -1+ (2x)/sqrt(9+x^2)`

At critical points (maxima, minima), the first drivative is zero.

`(df(x))/(dx) = 0`

`-1+ (2x)/sqrt(9+x^2) = 0`

`sqrt(9+x^2) = 2x`

by taking the square value of both sides,

`9+x^2 = 4x^2`

`3x^2 = 9`

`x^2 = 3`

`x = sqrt(3)`

Therefore at x = `sqrt(3)` , f(x) has a critical point.

C)To compute whether this is a minimum point, we have to calculate the second derivative of f(x),

`(d^2f(x))/dx^2 = (d((-1+(2x)/(sqrt(9+x^2))))/(dx))`

`(d^2f(x))/dx^2 = 0 + (2*sqrt(9+x^2)-2x*(1/2)*2x*(1/sqrt(9+x^2)))/(9+x^2)`

` ``(d^2f(x))/dx^2 = (2*(9+x^2)-2x^2*)/(9+x^2)^(3/2)`

`(d^2f(x))/dx^2 = 18/(9+x^2)^(3/2)`

The second derivative is positive for `sqrt(3)` (in fact for each and evry value). Therefore at x = `sqrt(3)`, f(x) has a minimum.

D) The minimum length of the cable is f(x) at x = `sqrt(3)`

`f(x) = (9-x)+2sqrt(9+x^2)`

`f(x) = (9-sqrt(3))+2sqrt(9+sqrt(3)^2)`

`f(x) = (9-sqrt(3))+2sqrt(9+3)`

`f(x) = (9-sqrt(3))+2sqrt(12)`

`f(x) = (9-sqrt(3))+2*2*sqrt(3)`

`f(x) = (9-sqrt(3))+4sqrt(3)`

`f(x) = 9+3sqrt(3)`

`f(x) = 14.196`

Therefore the minimum length of the cable is 14.196.