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 (12, 0) in the zy-plane, Springfield is at (0, 4), and Shelbyville is at (0,4). The cable runs from Centerville to some point (z, 0) on the-axis where it splits into two branches going to Springfield and Shelbyville. Find the location (z, 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 z:
f(x)=
We find that f(z) has a critical number at z =
To verify that f(x) has a minimum at this critical number we compute the second derivative f(z) and find that its value at the critical number is , a positive number.
Thus the minimum length of cable needed is
The Correct Answer and Explanation is :
To solve this problem, we are tasked with minimizing the total length of cable required to connect Centerville to Springfield and Shelbyville using a Y-shaped configuration. The cable connects Centerville at (12, 0) to a point (z, 0) on the x-axis, where it then branches out to Springfield at (0, 4) and Shelbyville at (0, 4). Our goal is to minimize the length of cable by finding the optimal value of z.
Step 1: Define the Total Length Function
The total length of cable consists of two parts:
- The distance from Centerville (12, 0) to the point (z, 0) on the x-axis.
- The distance from (z, 0) to Springfield (0, 4), and from (z, 0) to Shelbyville (0, 4).
For simplicity, the distance from Centerville to (z, 0) is the horizontal distance:
[
L_1(z) = |12 – z|
]
The distance from (z, 0) to Springfield (0, 4) is given by the Euclidean distance formula:
[
L_2(z) = \sqrt{(z – 0)^2 + (0 – 4)^2} = \sqrt{z^2 + 16}
]
Since the distance to Shelbyville is the same as the distance to Springfield, we can double this distance:
[
L_{\text{total}}(z) = L_1(z) + 2L_2(z) = |12 – z| + 2\sqrt{z^2 + 16}
]
Step 2: Minimize the Total Length Function
To find the value of ( z ) that minimizes the total length, we differentiate the function with respect to ( z ) and set the derivative equal to zero.
First, we differentiate ( L_1(z) = |12 – z| ) using the chain rule. Since ( |12 – z| = 12 – z ) when ( z \leq 12 ), we have:
[
\frac{d}{dz}L_1(z) = -1
]
Next, we differentiate ( L_2(z) = \sqrt{z^2 + 16} ):
[
\frac{d}{dz}L_2(z) = \frac{2z}{2\sqrt{z^2 + 16}} = \frac{z}{\sqrt{z^2 + 16}}
]
Thus, the derivative of ( L_{\text{total}}(z) ) is:
[
\frac{d}{dz}L_{\text{total}}(z) = -1 + 2 \cdot \frac{z}{\sqrt{z^2 + 16}}
]
To find the critical points, we set the derivative equal to zero:
[
-1 + 2 \cdot \frac{z}{\sqrt{z^2 + 16}} = 0
]
[
\frac{z}{\sqrt{z^2 + 16}} = \frac{1}{2}
]
Squaring both sides:
[
\frac{z^2}{z^2 + 16} = \frac{1}{4}
]
Multiply both sides by ( 4(z^2 + 16) ):
[
4z^2 = z^2 + 16
]
[
3z^2 = 16
]
[
z^2 = \frac{16}{3}
]
[
z = \frac{4}{\sqrt{3}} \approx 2.309
]
Step 3: Verify the Minimum
To verify that this is a minimum, we compute the second derivative of ( L_{\text{total}}(z) ):
[
\frac{d^2}{dz^2}L_{\text{total}}(z) = \frac{d}{dz}\left(-1 + 2 \cdot \frac{z}{\sqrt{z^2 + 16}}\right)
]
After performing the differentiation, we find that the second derivative is positive at ( z = \frac{4}{\sqrt{3}} ), confirming that this critical point corresponds to a minimum.
Step 4: Calculate the Minimum Length of Cable
Now, we substitute ( z = \frac{4}{\sqrt{3}} ) into the total length function to compute the minimum length of cable:
[
L_{\text{total}} \left(\frac{4}{\sqrt{3}}\right) = |12 – \frac{4}{\sqrt{3}}| + 2\sqrt{\left(\frac{4}{\sqrt{3}}\right)^2 + 16}
]
Simplifying the expression, we find the minimum total length of cable is approximately 22.52 units.
Thus, the optimal location to split the cable is at ( z = \frac{4}{\sqrt{3}} \approx 2.309 ), and the minimum amount of cable required is approximately 22.52 units.