The top one is the question, the middle one is how I solved it, and the bottom is how my professor solved it. I’m not sure why my answer is wrong and I’m confused about how my professor solved it. ### Optimization Problem: Minimizing Cost of Laying Pipe#### Problem DescriptionYou need to lay a pipe from point A to point B. The setup is as follows:- The direct horizontal distance between point A and point Q is 4 km.- The vertical distance from point Q to point B is 20 km.- The pipe can be laid underwater at a cost of $8,000 per kilometer or underground at a cost of $5,000 per kilometer.- The goal is to find the point P on line AQ that minimizes the total cost of laying the pipe from A to B.#### Diagram ExplanationA diagram illustrates the configuration with:- Point A located 4 km horizontally from point Q.- A vertical line from point Q to point B, which is 20 km long.- A proposed point P on line AQ, such that the distance from A to Q is labeled as x km.#### Cost Equation DerivationThe total cost $ C $ is given by the sum of the costs for laying the pipe underwater and underground:\[ C = 8000 \times \sqrt{x^2 + 16} + 5000 \times (20 - x)\]Where:- $ \sqrt{x^2 + 16} $ is the underwater distance (using the Pythagorean theorem).- $ 20 - x $ is the underground distance.#### Finding Minimum CostTo find the optimal point P that minimizes the cost:1. Differentiate the cost function $ C $ with respect to $ x $:\[\frac{dC}{dx} = 8000 \times \frac{x}{\sqrt{x^2 + 16}} - 5000 = 0 \]2. Solve for $ x $:\[\frac{8000x}{\sqrt{x^2 + 16}} = 5000\]3. Simplifying the equation, solve using algebra to find $ x $:\[8000x = 5000\sqrt{x^2 + 16}\]4. Solving this equation:\[64x^2 = 25(x^2 + 16)\]\[39x^2 = 400\]\[x^2 = \frac{400}{39}\]\[x \approx \sqrt{\frac{400}{39}} \approx 3.

Concept: The slope of a function can be found in its first derivative. The slope of the curve grows…

Answered: (5) It costs $8,000/km to lay pipe…

(5) It costs $8,000/km to lay pipe under the water and $5,000/km to walay pipe under the ground. Find the baut P that minimizes the cast of -4km- A

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

The top one is the question, the middle one is how I solved it, and the bottom is how my professor solved it. I’m not sure why my answer is wrong and I’m confused about how my professor solved it.

### Optimization Problem: Minimizing Cost of Laying Pipe

#### Problem Description

You need to lay a pipe from point A to point B. The setup is as follows:
- The direct horizontal distance between point A and point Q is 4 km.
- The vertical distance from point Q to point B is 20 km.
- The pipe can be laid underwater at a cost of $8,000 per kilometer or underground at a cost of $5,000 per kilometer.
- The goal is to find the point P on line AQ that minimizes the total cost of laying the pipe from A to B.

#### Diagram Explanation

A diagram illustrates the configuration with:
- Point A located 4 km horizontally from point Q.
- A vertical line from point Q to point B, which is 20 km long.
- A proposed point P on line AQ, such that the distance from A to Q is labeled as x km.

#### Cost Equation Derivation

The total cost $ C $ is given by the sum of the costs for laying the pipe underwater and underground:

\[
C = 8000 \times \sqrt{x^2 + 16} + 5000 \times (20 - x)
\]

Where:
- $ \sqrt{x^2 + 16} $ is the underwater distance (using the Pythagorean theorem).
- $ 20 - x $ is the underground distance.

#### Finding Minimum Cost

To find the optimal point P that minimizes the cost:
1. Differentiate the cost function $ C $ with respect to $ x $:

\[
\frac{dC}{dx} = 8000 \times \frac{x}{\sqrt{x^2 + 16}} - 5000 = 0
\]

2. Solve for $ x $:

\[
\frac{8000x}{\sqrt{x^2 + 16}} = 5000
\]

3. Simplifying the equation, solve using algebra to find $ x $:

\[
8000x = 5000\sqrt{x^2 + 16}
\]

4. Solving this equation:

\[
64x^2 = 25(x^2 + 16)
\]
\[
39x^2 = 400
\]
\[
x^2 = \frac{400}{39}
\]
\[
x \approx \sqrt{\frac{400}{39}} \approx 3.

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