How Can You Minimize the Installation Cost of Cable to an Island?

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SUMMARY

The discussion centers on minimizing the installation cost of electrical cable from a small island to the mainland, specifically addressing a scenario where the island is 80m from the shoreline and the nearest electrical connection is 200m along the shore. The cost to install cable across water is twice that of land. The optimal solution involves installing 92.4m of cable underwater and 153.8m along the shoreline, derived from the cost function C(x) = 2√(80²+x²) + (200-x) and its derivative C'(x) = (2x-1)/√(80²+x²).

PREREQUISITES
  • Understanding of basic calculus, specifically derivatives and optimization techniques.
  • Familiarity with the Pythagorean theorem for calculating distances in right triangles.
  • Knowledge of cost functions and how to derive them based on different installation scenarios.
  • Ability to interpret and analyze mathematical equations related to real-world applications.
NEXT STEPS
  • Study optimization problems in calculus to enhance problem-solving skills.
  • Learn about cost-benefit analysis in project management to apply similar principles in real-world scenarios.
  • Explore geometric applications of the Pythagorean theorem in engineering contexts.
  • Investigate different methods for minimizing costs in utility installation projects.
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Students in mathematics or engineering fields, project managers in utility installation, and anyone involved in cost optimization for infrastructure projects.

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Homework Statement


The owners of a small island want to bring in electricity from the mainland. The island is 80m from a straight shoreline at the closest point. The nearest electrical connection is 200m along the shore from that point. It costs twice as much to install cable across water than across land. What is the least expensive way to install the cable?

Homework Equations


C(x)= 2√(80^2+x^2) + (200-x)
C'(x)= (2x-1)/√(80^2+x^2)

The Attempt at a Solution


I drew a diagram of the scenario which created a right angle triangle. The height of it was 80 which I got from the info given and I said the base was "x" (the remaining length was 200-x since the entire point from the shoreline to the electrical connection was 200m). And then using the pythagorean theorem, I figured that the hypotenuse of the right angle triangle was √(80^2+x^2). And then, since it costs twice as much to install cable across water than across land, I knew that the first part of my cost equation was 2 x √(80^2+x^2) and then I just added (200-x) since that was what was left remaining to get to the electrical connection. I then found the derivative and got my zeros, making x=0.5 but that's apparently the wrong answer. The correct answer was 92.4m of cable below water and 153.8m along the shoreline.
 
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abizan said:

Homework Statement


The owners of a small island want to bring in electricity from the mainland. The island is 80m from a straight shoreline at the closest point. The nearest electrical connection is 200m along the shore from that point. It costs twice as much to install cable across water than across land. What is the least expensive way to install the cable?

Homework Equations


C(x)= 2√(80^2+x^2) + (200-x)
C'(x)= (2x-1)/√(80^2+x^2)

The Attempt at a Solution


I drew a diagram of the scenario which created a right angle triangle. The height of it was 80 which I got from the info given and I said the base was "x" (the remaining length was 200-x since the entire point from the shoreline to the electrical connection was 200m). And then using the pythagorean theorem, I figured that the hypotenuse of the right angle triangle was √(80^2+x^2). And then, since it costs twice as much to install cable across water than across land, I knew that the first part of my cost equation was 2 x √(80^2+x^2) and then I just added (200-x) since that was what was left remaining to get to the electrical connection. I then found the derivative and got my zeros, making x=0.5 but that's apparently the wrong answer. The correct answer was 92.4m of cable below water and 153.8m along the shoreline.

\frac{2x}{\sqrt{80^2+x^2}} -1 \neq \frac{2x-1}{\sqrt{80^2+x^2}}
 
Ray Vickson said:
\frac{2x}{\sqrt{80^2+x^2}} -1 \neq \frac{2x-1}{\sqrt{80^2+x^2}}
Ahh thank you, now it makes sense!
 

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