Respuesta :
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 ?
Answer:
A) [tex]\frac{2x}{\sqrt{3^{2}+{x^{2} } } } - 1[/tex]
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 = [tex]\sqrt{3^{2} +x^{2} }[/tex]
Shelbyville will have the same distance with Springfield because it is the same distance to the x axis as with Springfield
Dsh = [tex]\sqrt{(-3)^{2} +x^{2} }[/tex]
The total cable between 3 towns is the total cable calculated above:
D = Dc + Dsp + Dsh
D(x) = 11 - x + 2[tex]\sqrt{(3)^{2} +x^{2} }[/tex]
= [tex]2\sqrt{3^{2}+ x^2} +11 - x[/tex]
= [tex]2(3^{2}+ x^{2})^{1/2} + 11 - x[/tex]
a) Minimize the function
Do differentiation to find the first derivative of the above function:
D'(x) = dD(x)/dx
= 2(1/2)[tex](3^{2} +x^{2} )^{1/2} (2x)[/tex] - 1
= [tex]\frac{2x}{\sqrt{3^{2}+{x^{2} } } } - 1[/tex]
b) Critical number
To find the critical of function, make D'(x) = 0
[tex]\frac{2x}{\sqrt{3^{2}+{x^{2} } } } - 1[/tex] = 0
[tex]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\\[/tex]
Critical number at x:
[tex]\sqrt{3} x - 3=0\\x = \frac{3}{\sqrt{3} } \\[/tex]
x = 1.732
[tex]\sqrt{3} x + 3=0\\x = \frac{-3}{\sqrt{3} } \\[/tex]
x = -1.732
Since x couldn't be negative,
[tex]\sqrt{3} x - 3=0\\x = \frac{3}{\sqrt{3} } \\[/tex]
x = 1.732
c) 2nd derivative
Differentiate for 2nd derivative (can use composite method)
D''(x) = ddD(x)/dx
= [tex]\frac{18}{(x^{2} +9)^{3/2} }[/tex]
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) = [tex]2(3^{2}+ x^{2})^{1/2} + 11 - x[/tex]
We get D(x) = 16.2
