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(500-X) 20M 500M

(500-X) 20M 500M

Calculus: Early Transcendentals
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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I already have most of this done. I just need help filling in the blanks. Calculus 1. (Optimization) :-)
### Educational Content on Optimization Problem #### Diagram Description The diagram illustrates a river-crossing scenario where a path from one point on the riverbank to another utilizes both walking on the land and swimming across the river. The path forms a right triangle with the legs labeled as \(x\) and \(20 \, \text{m}\) (river width), and the hypotenuse labeled as \(500-x\), which represents the remaining land distance. The total horizontal distance is \(500 \, \text{m}\). #### Problem Setup We're given: - Land distance, \(x\), and swimming distance, \((500-x)\). - Width of the river is \(20 \, \text{m}\). #### Objective Minimize \(D(x)\), which is the total distance function: \[ D = x + 5y \] 1. **Express \(y\) in terms of \(x\):** \[ y = \sqrt{20^2 + (500-x)^2} = \sqrt{400 + 250000 - 1000x + x^2} = \sqrt{250400 - 1000x + x^2} \] 2. **Define the Distance Function:** \[ D(x) = x + 5y = x + 5\left(\sqrt{250400 - 1000x + x^2}\right) \] Domain: \( x \in [0, 500] \) 3. **Differentiate \(D(x)\):** \[ \frac{d}{dx} D(x) = 1 + 5 \left(\frac{1}{2} \right) \left(2x - 1000\right) \left(\frac{1}{\sqrt{250400 - 1000x + x^2}}\right) \] Simplifying: \[ \frac{d}{dx} D(x) = 1 + \frac{5(2x-1000)}{2\sqrt{250400 - 1000x + x^2}} \] 4. **Set the Derivative to Zero for Minimization:** \[ 25(4x^2-4000x + 10000) = 4(250400 - 1000x + x^2) \
Transcribed Image Text:### Educational Content on Optimization Problem #### Diagram Description The diagram illustrates a river-crossing scenario where a path from one point on the riverbank to another utilizes both walking on the land and swimming across the river. The path forms a right triangle with the legs labeled as \(x\) and \(20 \, \text{m}\) (river width), and the hypotenuse labeled as \(500-x\), which represents the remaining land distance. The total horizontal distance is \(500 \, \text{m}\). #### Problem Setup We're given: - Land distance, \(x\), and swimming distance, \((500-x)\). - Width of the river is \(20 \, \text{m}\). #### Objective Minimize \(D(x)\), which is the total distance function: \[ D = x + 5y \] 1. **Express \(y\) in terms of \(x\):** \[ y = \sqrt{20^2 + (500-x)^2} = \sqrt{400 + 250000 - 1000x + x^2} = \sqrt{250400 - 1000x + x^2} \] 2. **Define the Distance Function:** \[ D(x) = x + 5y = x + 5\left(\sqrt{250400 - 1000x + x^2}\right) \] Domain: \( x \in [0, 500] \) 3. **Differentiate \(D(x)\):** \[ \frac{d}{dx} D(x) = 1 + 5 \left(\frac{1}{2} \right) \left(2x - 1000\right) \left(\frac{1}{\sqrt{250400 - 1000x + x^2}}\right) \] Simplifying: \[ \frac{d}{dx} D(x) = 1 + \frac{5(2x-1000)}{2\sqrt{250400 - 1000x + x^2}} \] 4. **Set the Derivative to Zero for Minimization:** \[ 25(4x^2-4000x + 10000) = 4(250400 - 1000x + x^2) \
**Problem Statement:** A person on one side of a straight river that is 20 meters wide needs to reach the other side of the river at a point 500 meters upstream as fast as possible. If she can run 5 times faster than she can swim, how far along the shore should she run before jumping into the river and swimming diagonally the rest of the way to her target if she wants to get to the point upstream as fast as possible? **Analysis:** To solve this problem, we must consider the time it takes to both run and swim. The person has two options: 1. **Run along the shore** a certain distance, \( x \), and then swim diagonally across the river to the target point. 2. **Swim directly** across the river without any running. The objective is to minimize the total travel time. - The width of the river is 20 meters. - The target point is 500 meters upstream. - Running speed is 5 times her swimming speed. We can denote: - \( v \) as her swimming speed. - \( 5v \) as her running speed. To find the optimal distance \( x \) she should run, one can derive it by minimizing the total time equation which involves both running and swimming components. This is a classical optimization problem considering velocity and path constraints.
Transcribed Image Text:**Problem Statement:** A person on one side of a straight river that is 20 meters wide needs to reach the other side of the river at a point 500 meters upstream as fast as possible. If she can run 5 times faster than she can swim, how far along the shore should she run before jumping into the river and swimming diagonally the rest of the way to her target if she wants to get to the point upstream as fast as possible? **Analysis:** To solve this problem, we must consider the time it takes to both run and swim. The person has two options: 1. **Run along the shore** a certain distance, \( x \), and then swim diagonally across the river to the target point. 2. **Swim directly** across the river without any running. The objective is to minimize the total travel time. - The width of the river is 20 meters. - The target point is 500 meters upstream. - Running speed is 5 times her swimming speed. We can denote: - \( v \) as her swimming speed. - \( 5v \) as her running speed. To find the optimal distance \( x \) she should run, one can derive it by minimizing the total time equation which involves both running and swimming components. This is a classical optimization problem considering velocity and path constraints.
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