Math / Calculus

QuestionCenterville 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 $(8,0)$ 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 $x$ : $f(x)=$ $\square$ We find that $f(x)$ has a critical number at $x=$ $\square$ To verify that $f(x)$ has a minimum at this critical number we compute the second derivative $f^{\prime \prime}(x)$ and find that its value at the critical number is $\square$ , a positive number.
Thus the minimum length of cable needed is $\square$ Question Help: $\square$ Video Submit Question

Studdy Solution

STEP 1

What is this asking? Find the best spot on the x-axis to put a cable splitter so that the total cable length from Centerville to Springfield and Shelbyville is as short as possible! Watch out! Don't forget to check that your answer actually minimizes the cable length, not maximizes it!

STEP 2

1. Define the function
2. Optimize the formula
3. Calculate the result
4. Verify minimum

STEP 3

We need a formula for the total cable length.
Let's call the point where the cable splits $(x, 0)$ .
The distance from Centerville $(8, 0)$ to the splitting point $(x, 0)$ is simply $8 - x$ .
We're subtracting $x$ from 8 since the splitting point is to the left of Centerville.

STEP 4

The distance from the splitting point $(x, 0)$ to Springfield $(0, 5)$ can be found using the Pythagorean theorem!
We have a right triangle with legs of length $x$ and $5$ , so the distance is $\sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}$ .

STEP 5

Similarly, the distance from the splitting point $(x, 0)$ to Shelbyville $(0, -5)$ is also $\sqrt{x^2 + (-5)^2} = \sqrt{x^2 + 25}$ .
Notice that the distances to Springfield and Shelbyville are the same due to symmetry!

STEP 6

So, the total cable length, which we'll call $f(x)$ , is the sum of these three distances: $f(x) = (8 - x) + \sqrt{x^2 + 25} + \sqrt{x^2 + 25} = 8 - x + 2\sqrt{x^2 + 25}.$

STEP 7

To minimize $f(x)$ , we need to find its critical points!
Let's take the derivative of $f(x)$ with respect to $x$ : $f'(x) = -1 + 2 \cdot \frac{1}{2\sqrt{x^2 + 25}} \cdot 2x = -1 + \frac{2x}{\sqrt{x^2 + 25}}.$ Remember, the derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$ by the chain rule.

STEP 8

Now, set $f'(x)$ equal to zero and solve for $x$ : $-1 + \frac{2x}{\sqrt{x^2 + 25}} = 0.$ $\frac{2x}{\sqrt{x^2 + 25}} = 1.$ $2x = \sqrt{x^2 + 25}.$ $4x^2 = x^2 + 25.$ $3x^2 = 25.$ $x^2 = \frac{25}{3}.$ $x = \pm\frac{5}{\sqrt{3}}.$ Since the splitting point must be between Centerville and the y-axis, we take the positive solution: $x = \frac{5}{\sqrt{3}}$ .

STEP 9

Plug this value of $x$ back into our formula for $f(x)$ to find the minimum cable length: \begin{align} f\left(\frac{5}{\sqrt{3}}\right) &= 8 - \frac{5}{\sqrt{3}} + 2\sqrt{\left(\frac{5}{\sqrt{3}}\right)^2 + 25} \\ &= 8 - \frac{5}{\sqrt{3}} + 2\sqrt{\frac{25}{3} + 25} \\ &= 8 - \frac{5}{\sqrt{3}} + 2\sqrt{\frac{100}{3}} \\ &= 8 - \frac{5}{\sqrt{3}} + 2 \cdot \frac{10}{\sqrt{3}} \\ &= 8 + \frac{15}{\sqrt{3}} \\ &= 8 + \frac{15\sqrt{3}}{3} = 8 + 5\sqrt{3}.\end{align}

STEP 10

The second derivative of $f(x)$ is: $f''(x) = \frac{2\sqrt{x^2 + 25} - 2x(\frac{1}{2}(x^2+25)^{-1/2}(2x))}{x^2 + 25} = \frac{2(x^2 + 25) - 2x^2}{(x^2 + 25)^{3/2}} = \frac{50}{(x^2 + 25)^{3/2}}.$ Since $f''(x)$ is always positive, our critical point corresponds to a minimum!

STEP 11

The location that minimizes the cable length is $(\frac{5}{\sqrt{3}}, 0)$ , and the minimum length of cable needed is $8 + 5\sqrt{3}$ .

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Studdy Solution

STEP 1

What is this asking? Find the best spot on the x-axis to put a cable splitter so that the total cable length from Centerville to Springfield and Shelbyville is as short as possible! Watch out! Don't forget to check that your answer actually minimizes the cable length, not maximizes it!

STEP 2

1. Define the function
2. Optimize the formula
3. Calculate the result
4. Verify minimum

STEP 3

We need a formula for the total cable length.
Let's call the point where the cable splits $(x, 0)$ .
The distance from Centerville $(8, 0)$ to the splitting point $(x, 0)$ is simply $8 - x$ .
We're subtracting $x$ from 8 since the splitting point is to the left of Centerville.

STEP 4

The distance from the splitting point $(x, 0)$ to Springfield $(0, 5)$ can be found using the Pythagorean theorem!
We have a right triangle with legs of length $x$ and $5$ , so the distance is $\sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}$ .

STEP 5

Similarly, the distance from the splitting point $(x, 0)$ to Shelbyville $(0, -5)$ is also $\sqrt{x^2 + (-5)^2} = \sqrt{x^2 + 25}$ .
Notice that the distances to Springfield and Shelbyville are the same due to symmetry!

STEP 6

So, the total cable length, which we'll call $f(x)$ , is the sum of these three distances: $f(x) = (8 - x) + \sqrt{x^2 + 25} + \sqrt{x^2 + 25} = 8 - x + 2\sqrt{x^2 + 25}.$

STEP 7

STEP 8

STEP 9

STEP 10

The second derivative of $f(x)$ is: $f''(x) = \frac{2\sqrt{x^2 + 25} - 2x(\frac{1}{2}(x^2+25)^{-1/2}(2x))}{x^2 + 25} = \frac{2(x^2 + 25) - 2x^2}{(x^2 + 25)^{3/2}} = \frac{50}{(x^2 + 25)^{3/2}}.$ Since $f''(x)$ is always positive, our critical point corresponds to a minimum!

STEP 11

The location that minimizes the cable length is $(\frac{5}{\sqrt{3}}, 0)$ , and the minimum length of cable needed is $8 + 5\sqrt{3}$ .

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Studdy Solution

STEP 1

What is this asking? Find the best spot on the x-axis to put a cable splitter so that the total cable length from Centerville to Springfield and Shelbyville is as short as possible! Watch out! Don't forget to check that your answer actually *minimizes* the cable length, not maximizes it!

STEP 2

1. Define the function2. Optimize the formula3. Calculate the result4. Verify minimum

STEP 3

STEP 4

The distance from the splitting point (x,0)(x, 0)(x,0) to Springfield (0,5)(0, 5)(0,5) can be found using the Pythagorean theorem!We have a right triangle with legs of length xxx and 555, so the distance is x2+52=x2+25\sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}x2+52​=x2+25​.

STEP 5

Similarly, the distance from the splitting point (x,0)(x, 0)(x,0) to Shelbyville (0,−5)(0, -5)(0,−5) is also x2+(−5)2=x2+25\sqrt{x^2 + (-5)^2} = \sqrt{x^2 + 25}x2+(−5)2​=x2+25​.Notice that the distances to Springfield and Shelbyville are the same due to symmetry!

STEP 6

So, the total cable length, which we'll call f(x)f(x)f(x), is the sum of these three distances: f(x)=(8−x)+x2+25+x2+25=8−x+2x2+25.f(x) = (8 - x) + \sqrt{x^2 + 25} + \sqrt{x^2 + 25} = 8 - x + 2\sqrt{x^2 + 25}.f(x)=(8−x)+x2+25​+x2+25​=8−x+2x2+25​.

STEP 7

STEP 8

STEP 9

STEP 10

STEP 11

The location that minimizes the cable length is (53,0)(\frac{5}{\sqrt{3}}, 0)(3​5​,0), and the minimum length of cable needed is 8+538 + 5\sqrt{3}8+53​.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? Find the best spot on the x-axis to put a cable splitter so that the total cable length from Centerville to Springfield and Shelbyville is as short as possible! Watch out! Don't forget to check that your answer actually minimizes the cable length, not maximizes it!

1. Define the function
2. Optimize the formula
3. Calculate the result
4. Verify minimum

The distance from the splitting point $(x, 0)$ to Springfield $(0, 5)$ can be found using the Pythagorean theorem!
We have a right triangle with legs of length $x$ and $5$ , so the distance is $\sqrt{x^2 + 5^2} = \sqrt{x^2 + 25}$ .

Similarly, the distance from the splitting point $(x, 0)$ to Shelbyville $(0, -5)$ is also $\sqrt{x^2 + (-5)^2} = \sqrt{x^2 + 25}$ .
Notice that the distances to Springfield and Shelbyville are the same due to symmetry!

So, the total cable length, which we'll call $f(x)$ , is the sum of these three distances: $f(x) = (8 - x) + \sqrt{x^2 + 25} + \sqrt{x^2 + 25} = 8 - x + 2\sqrt{x^2 + 25}.$

The location that minimizes the cable length is $(\frac{5}{\sqrt{3}}, 0)$ , and the minimum length of cable needed is $8 + 5\sqrt{3}$ .