Optimization Problem: Minimizing Cost

[alli.b.]

Question:
An isolated farm house is located on the bank of a straight canal that is 200 m wide. Electric power for the house must come from a power station on the opposite side of the canal, 500 m downstream. If it cost twice as much to lay cable underwater as it does to lay cable on land, what path should be chosen to minimize the cost?

I am currently stuck in the creating an equation step, and would like a little assistance in getting started. The part that is confusing me is creating an equation without the actual costs. How would I write it not given the costs, but knowing one cost is double the other.

I do not have any work to show, as it is one of the first steps, but I will include the diagram. I will show the other question I am following as a reference.

Diagram. Page 1 of reference, page 2 of reference.

 

[skeeter]

underwater cable distance, [MATH]\sqrt{200^2+x^2}[/MATH]
land distance, [MATH]500-x[/MATH]
you can consider the price of 1 meter of cable as 1 unit of ?

ergo, cost as a function of distance x is

[MATH]C = 2\sqrt{200^2+x^2} + (500-x)[/MATH]
determine [MATH]\frac{dC}{dx}[/MATH] and minimize ...

 

[Cubist]

you can consider the price of 1 meter of cable as 1 unit of ?

...this is a good strategy, but I would recommend clarifying that the above is the "on land cost"...

you can consider the price of 1 meter of cable on land as 1 unit of ?

--

@alli.b. another approach (not better than the above, just different) is to say...

Let z be the cost of laying 1m of cable on land.
Therefore the cost of laying 1m of cable underwater is 2z.
Obviously, treat z as a constant while differentiating.

 

[alli.b.]

Cubist,
I tried both strategies, I did have more luck with just using units. I wasn't sure how to proceed with using the z any further past finding the second derivative.
Do you have any suggestions for how I would continue to finish the question with the using z method? Or does just using units and having the note about what I am doing in the Establish Variables section make it easier to understand/follow?

Thank you for your help!


Below is my attempt of using z, I got stuck when confirming max/min.



Below is my attempt just using 1 as 1 unit/m of cable on land and 2 units/m of cable underwater.

 

[skeeter]

[MATH]C = 2z\sqrt{200^2+x^2} + z(500-x)[/MATH]
[MATH]C = z\bigg[2\sqrt{200^2+x^2} + (500-x)\bigg][/MATH]
[MATH]\frac{dC}{dx} = z\bigg[\frac{2x}{\sqrt{200^2+x^2}} - 1 \bigg][/MATH]
setting the derivative equal to 0 ...

[MATH]2x = \sqrt{200^2+x^2}[/MATH]
[MATH]4x^2 = 200^2+x^2 \implies x = \frac{200}{\sqrt{3}}[/MATH]
[MATH]\frac{d^2C}{dx^2} = z\bigg[\frac{2\sqrt{200^2+x^2} - 2x \cdot \frac{x}{\sqrt{200^2+x^2}}}{200^2+x^2}\bigg][/MATH]
multiply numerator & denominator by the radical ...

[MATH]\frac{d^2C}{dx^2} = z\bigg[\frac{2(200^2+x^2) - 2x^2}{(200^2+x^2)^{3/2}}\bigg][/MATH]
[MATH]\frac{d^2C}{dx^2} = z\bigg[\frac{2 \cdot 200^2}{(200^2+x^2)^{3/2}}\bigg] > 0 \text{ for all }x>0[/MATH]
therefore, [math]C(x)[/math] is concave up for all [math]x > 0 \implies C\left(\frac{200}{\sqrt{3}}\right)[/math] is a minimum.

... and it's clear that [MATH]z[/MATH] doesn't really matter.

 

[Cubist]

@alli.b. please see attached for a mistake that you made in your work...



I recommend that you simplify as you go, on your second or third line above you could divide both sides of the equation by z to eliminate z entirely! Try to spot opportunities like this.

Also note in @skeeter 's post#5 that factoring is better than expanding in this situation. Don't be too quick to expand (you might call it distribute).

BUT your work is nicely laid out at a high level, and you have showed us your thinking. Very well done for that!

 

[alli.b.]

@skeeter @Cubist

Thank you both so much for your help!