1. ## Stuck on question

There is a question that says,

A new cottage s built across the river and 300m downstream from the nearest telephone relay station. The river is 120m wide. In order to wire the cottage for phone service, wire will be laid across the river under water, and along the edge of the river above ground. The cost to lay wire under water is $15/m and the cost to lay wire above ground is$10/m. How much wire should be laid under water to minimize the cost? Be sure to check that your answer is indeed a minimum.

I drew a picture to illustrate the cottage, river and telephone station,

I did a calculation trying to see what the total cost would be if the wire was to be laid by going across the river vertically so the whole 120m and then 300m horizontally to the cottage,

C=15(120)+10(300) \par
=4800

and then I decided to see how it would look if the wire went diagonal through the river to the other side by doing the pythagorean theorem pretending that the 120m across the river was one side of a square and then another side would be 120m and the diagonal after using the pythagorean was 169.7m. Then I went 300-169.7=130.3 and did this,

C=15(169.7)+10(130.3)
=3848.5

So it would cost less to go across the river diagonally, but I need to figure this out using a function. So I know that the function should look like this,

c^2=a^2+b^2
c^2=x^2+120^2
sqrt(c^2)=sqrt(x^2+14400)
c=sqrt(x^2+14400)

C=10(300-x)+15(x^2+14400)^(1/2)
C'=-10+15x(x^2+14400)^-(1/2)

So now I need to find the x's and this is where I am confused.

0=-10+15x(x^2+14400)^-(1/2)

I was struggling to figure out how to do this so I found someone figured this out online and came up with the answer,

1)
C=10(300-x) +15(x^2+120^2)^(1/2)

dC/dx= 15x/(x^2+14400)^(1/2) -10
0=15x/(x^2+14400)^(1/2) -10

no steps showing how they arrived at this,
x=107.33

161 m of wire under water

2)

y=(2000+200x)(48-2x)

dy/dx=5600-800x
0=5600-800x
x=7

$34 produces max profit And then I tried plugging, 0=-10+15x(x^2+14400)^-(1/2) into mathway.com and they show the x's as 0, 120i, 120i. So once again I'm at a loss as to which was is right, although mathway provides steps so I'm more inclined to go with that answer however I'm not even sure how to solve past that answer either. I'm teaching myself calculus and I apologize if I'm making this out to be more complicating than it probably is but I want to make sure I can figure these problems out the right way. Thanks for your time and help! Edit: don't mind the attachment at the bottom of my post, I'm not sure how to delete it. 2. ## Re: Stuck on question A new cottage s built across the river and 300m downstream from the nearest telephone relay station. The river is 120m wide. In order to wire the cottage for phone service, wire will be laid across the river under water, and along the edge of the river above ground. The cost to lay wire under water is 15 dollars per meter and the cost to lay wire above ground is10 dollars per meter. How much wire should be laid under water to minimize the cost? Be sure to check that your answer is indeed a minimum. Cost = (15 dollars)(water distance) + (10 dollars)(land distance) let$x$= distance downstream from the cottage to start the land wire$\displaystyle C(x) = 15\sqrt{120^2+x^2} + 10(300-x)\displaystyle \frac{dC}{dx} = \frac{15x}{\sqrt{120^2+x^2}} - 10$set$\displaystyle \frac{dC}{dx} = 0$...$\displaystyle \frac{15x}{\sqrt{120^2+x^2}} = 10\displaystyle 3x = 2\sqrt{120^2+x^2}\displaystyle 9x^2 = 4(120^2+x^2)\displaystyle 5x^2 = 4 \cdot 120^2\displaystyle x = 48\sqrt{5} \approx 107.33 \, \text{meters}$water distance =$\displaystyle \sqrt{120^2+5 \cdot 48^2} \approx 161 \, \text{meters}\$

Now ... how would you confirm that this distance yields the minimum cost?

3. ## Re: Stuck on question

Thanks for the reply, I'm just curious how do you get from,

\frac{15x}{\sqrt{120^2+x^2}} = 10

to,

3x = 2\sqrt{120^2+x^2}

like what steps did you do?

4. ## Re: Stuck on question

Oh I get it now, you divided 15/5 = 3 and 10/5 = 2. Then to get rid of the radical you squared both sides of the equation. I had to figure out the steps but I can understand now. Thanks again!

5. ## Re: Stuck on question

Just got to this question and I'm lost

Sent from my SGH-I747M using Tapatalk