Mathematics, 14.12.2019 06:31 theannakittler7327

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. centerville is located at (11,0) in the xy-plane, springfield is at (0,3), and shelbyville is at (0,-3). to save on the cost of cable, greedy cablevision wants to arrange the cable in a y-shaped configuation, running cable from centerville to some point (x,0) on the x-axis where it then splits into two branches, one going to springfield and one to 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 conclusion by answering the following questions. (a) to solve this problem we need to minimize the following function of x:

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Answer from: Halessoftball

There is some missing part of this question. The following should be the complete 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 configuation. Centerville is located at(11,0) in the xy-plane, Springfield is at (0,3), and Shelbyville is at (0,-3). The cable runs from Centerville to some point (x,0) on the -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.

A) To solve this problem we need to minimize the following function of x:

f(x)=?

B) We find that f(x) has a critical number at x=?

C) 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.

D) Thus the minimum length of cable needed is ?

A) $\frac{2x}{\sqrt{3^{2}+{x^{2} } } } - 1$

B) x = 1.732 and x = -1.732

C) 0.433 (positive number)

D) 16.2

Step-by-step explanation:

First note that there is a point of where the cable from Centerville split (x,0).

The distance from Centreville, Dc

Dc = 11 - x

While distance from Springfield is (by using theorem phytagoras)

Dsp = $\sqrt{3^{2} +x^{2} }$

Shelbyville will have the same distance with Springfield because it is the same distance to the x axis as with Springfield

Dsh = $\sqrt{(-3)^{2} +x^{2} }$

The total cable between 3 towns is the total cable calculated above:

D = Dc + Dsp + Dsh

D(x) = 11 - x + 2 $\sqrt{(3)^{2} +x^{2} }$

= $2\sqrt{3^{2}+ x^2} +11 - x$

= $2(3^{2}+ x^{2})^{1/2} + 11 - x$

a) Minimize the function

Do differentiation to find the first derivative of the above function:

D'(x) = dD(x)/dx

= 2(1/2) $(3^{2} +x^{2} )^{1/2} (2x)$ - 1

= $\frac{2x}{\sqrt{3^{2}+{x^{2} } } } - 1$

b) Critical number

To find the critical of function, make D'(x) = 0

$\frac{2x}{\sqrt{3^{2}+{x^{2} } } } - 1$ = 0

$2x = \sqrt{3^{2}+x^{2} } \\(2x)^{2} =3^{2}+ x^{2}\\4x^{2} = 3^{2}+ x^{2}\\3x^{2} = 9\\3x^{2} - 9 = 0\\(\sqrt{3}x)^{2}-3^{2}=0\\(\sqrt{3}x +3)(\sqrt{3}x -3)=0\\$

Critical number at x:

$\sqrt{3} x - 3=0\\x = \frac{3}{\sqrt{3} } \\$

x = 1.732

$\sqrt{3} x + 3=0\\x = \frac{-3}{\sqrt{3} } \\$

x = -1.732

Since x couldn't be negative,

$\sqrt{3} x - 3=0\\x = \frac{3}{\sqrt{3} } \\$

x = 1.732

c) 2nd derivative

Differentiate for 2nd derivative (can use composite method)

D''(x) = ddD(x)/dx

= $\frac{18}{(x^{2} +9)^{3/2} }$

subs. x = 1.732, D''(x) = 0.433

subs. x = -1.732, D''(x) = 0.433

Positive number -> hence we have minimum value.

We'll get the minimum cable length by substituing x into the original equation.

Substitute x = 1.732 into D(x) = $2(3^{2}+ x^{2})^{1/2} + 11 - x$

We get D(x) = 16.2

Answer from: Quest

7for nonmembers i think

Answer from: Quest

step-by-step explanation:

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