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 the cable by arranging the cable in a Y-shaped configuation. Centerville is located at ( 10 , 0 ) in the x y -plane, Springfield is at ( 0 , 5 ) , and Shelbyville is at ( 0 , − 5 ) . 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

Respuesta :

Answer:

hi i dont understand this

Step-by-step explanation:

sorry

The function will be

[tex]f(x) = 9 - x + \sqrt[2]{ {x}^{2} + 9} [/tex]

and minimum value required will be

[tex]9 + \sqrt[3]{3} [/tex]

How to solve this problem?

The steps are as follow:

[tex]f(x) = (9 - x) + \sqrt[2]{ (x^2 + 3^2) } \\ =(9 - x) + \sqrt[2]{ (x^2 +9)} \\ f(x) = \frac{2x}{ \sqrt{ {x}^{2} + 9} } - 1 \\ f(x) = 0 \\ \frac{2x}{ \sqrt{ {x}^{2} + 9} } - 1 = 0 \\ \frac{2x}{ \sqrt{ {x}^{2} + 9}} = 1 \\ 2x = \sqrt{ {x}^{2} + 9} \\ 4{x}^{2} = { x}^{2} + 9 \\ 3 {x}^{2} = 9 \\ {x} = \sqrt{3} [/tex]

so putting value of in f(x) we will get

[tex]f( \sqrt{3} ) = 9 - \sqrt{3} + \sqrt[2]{ { (\sqrt{3} )}^{2} + 9} \\ = 9 + 2 \sqrt{3} [/tex]

So The function will be

[tex]f(x) = 9 + \sqrt[2]{ {x}^{2} + 9}[/tex]

and minimum value required will be

[tex]9 + \sqrt[3]{3} [/tex]

Learn more about functions here:

https://brainly.com/question/2833285

#SPJ2