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

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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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