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 in the -plane, Springfield is at , and Shelbyville is at . The
cable runs from Centerville to some point on the -axis where it splits into two branches going to
Springfield and Shelbyville. Find the location 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 :

We find that has a critical number at
To verify that has a minimum at this critical number we compute the second derivative and
find that its value at the critical number is

Ask by Hodgson Norris. in the United States

Mar 23,2025

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 in the -plane, Springfield is at , and Shelbyville is at . The
cable runs from Centerville to some point on the -axis where it splits into two branches going to
Springfield and Shelbyville. Find the location 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 :

We find that has a critical number at
To verify that has a minimum at this critical number we compute the second derivative and
find that its value at the critical number is

Ask by Hodgson Norris. in the United States

Mar 23,2025

UpStudy AI · Accepted Answer

The length of cable needed is made up of three segments: one from Centerville at $$ (9,0) $$ to the junction point $$ (x,0) $$ on the $$ x $$-axis, and two from $$ (x,0) $$ to the towns Springfield at $$ (0,6) $$ and Shelbyville at $$ (0,-6) $$. We begin by expressing the total length as a function of $$ x $$.

### Step 1. Write the Function

The distance from Centerville to the junction is
$$
|9 - x|.
$$

The distance from $$ (x,0) $$ to Springfield (or Shelbyville) is
$$
\sqrt{(x-0)^2 + (0-6)^2} = \sqrt{x^2 + 36}.
$$

Because there are two identical branches (one to Springfield and another to Shelbyville), the total cable length is
$$
f(x) = |9-x| + 2\sqrt{x^2+36}.
$$

Since we expect the optimal $$ x $$ to lie between 0 and 9 (making $$9-x$$ positive), we have
$$
f(x) = 9 - x + 2\sqrt{x^2+36}.
$$

Thus, the function to minimize is
$$
\boxed{f(x)=9-x+2\sqrt{x^2+36}}.
$$

### Step 2. Find the Critical Number

To minimize $$ f(x) $$, we compute its first derivative. Differentiate term by term:
$$
\frac{d}{dx}(9-x) = -1,
$$
and
$$
\frac{d}{dx}\sqrt{x^2+36} = \frac{x}{\sqrt{x^2+36}}.
$$
Thus, the derivative of $$ f(x) $$ is
$$
f'(x) = -1 + 2\cdot\frac{x}{\sqrt{x^2+36}}.
$$

Setting $$ f'(x) = 0 $$, we have
$$
-1 + \frac{2x}{\sqrt{x^2+36}}=0.
$$
Solve for $$ x $$:
$$
\frac{2x}{\sqrt{x^2+36}} = 1.
$$
Multiply both sides by $$ \sqrt{x^2+36} $$:
$$
2x = \sqrt{x^2+36}.
$$
Square both sides to eliminate the square root:
$$
4x^2 = x^2 + 36.
$$
Subtract $$ x^2 $$ from both sides:
$$
3x^2 = 36,
$$
so
$$
x^2 = 12.
$$
Since $$ x>0 $$, we have
$$
x = 2\sqrt{3}.
$$
Thus, the critical number is
$$
\boxed{x=2\sqrt{3}}.
$$

### Step 3. Verify it is a Minimum via the Second Derivative Test

Differentiate $$ f'(x) $$:
$$
f'(x) = -1 + 2\frac{x}{\sqrt{x^2+36}} = -1 + 2x(x^2+36)^{-1/2}.
$$

Differentiate the term $$ 2x(x^2+36)^{-1/2} $$ using the product rule:
$$
\frac{d}{dx}\left(2x(x^2+36)^{-1/2}\right)
= 2(x^2+36)^{-1/2} + 2x\left(-\frac{1}{2}\right)(x^2+36)^{-3/2}(2x).
$$
Simplify the second term:
$$
2x\left(-\frac{1}{2}\right)(2x)(x^2+36)^{-3/2} = -\frac{2x^2}{(x^2+36)^{3/2}}.
$$
Thus,
$$
f''(x) = 2(x^2+36)^{-1/2} - \frac{2x^2}{(x^2+36)^{3/2}}.
$$
Factor out $$ \frac{2}{(x^2+36)^{3/2}} $$:
$$
f''(x) = \frac{2}{(x^2+36)^{3/2}}\left[(x^2+36) - x^2\right]
= \frac{2\cdot 36}{(x^2+36)^{3/2}}
= \frac{72}{(x^2+36)^{3/2}}.
$$

Now evaluate at $$ x=2\sqrt{3} $$. Notice
$$
x^2+36 = 12+36 = 48.
$$
So,
$$
f''(2\sqrt{3}) = \frac{72}{48^{3/2}}.
$$
Calculate $$ 48^{3/2} $$:
$$
48^{3/2} = (\sqrt{48})^3 = (4\sqrt{3})^3 = 64\cdot3\sqrt{3} = 192\sqrt{3}.
$$
Thus,
$$
f''(2\sqrt{3}) = \frac{72}{192\sqrt{3}} = \frac{3}{8\sqrt{3}},
$$
which is positive. Since $$ f''(2\sqrt{3}) > 0 $$, the function has a local minimum at $$ x=2\sqrt{3} $$.

### Step 4. Compute the Minimum Cable Length

Substitute $$ x=2\sqrt{3} $$ into $$ f(x) $$:
$$
f(2\sqrt{3}) = 9 - 2\sqrt{3} + 2\sqrt{(2\sqrt{3})^2+36}.
$$
Compute inside the square root:
$$
(2\sqrt{3})^2 = 12, \quad \text{so} \quad \sqrt{12+36} = \sqrt{48} = 4\sqrt{3}.
$$
Thus,
$$
f(2\sqrt{3}) = 9 - 2\sqrt{3} + 2(4\sqrt{3}) = 9 - 2\sqrt{3} + 8\sqrt{3} = 9 + 6\sqrt{3}.
$$

### Final Answers

1. The function minimized is
$$
\boxed{f(x)=9-x+2\sqrt{x^2+36}}.
$$
2. The function has a critical number at
$$
\boxed{x=2\sqrt{3}}.
$$
3. The second derivative at $$ x=2\sqrt{3} $$ is
$$
\boxed{f''(2\sqrt{3})=\frac{3}{8\sqrt{3}} > 0}.
$$
4. The minimum amount of cable required is
$$
\boxed{9+6\sqrt{3}}.
$$
To minimize the cable length, the junction point should be at $$ x = 2\sqrt{3} $$. The total cable needed is $$ 9 + 6\sqrt{3} $$ units.

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