[MOVED] Finding formula for cheapest path for pipe

[Naur]

I'm not exactly sure, as I just don't know how to do it, but it probably concerns finding a formula to describe cost, then finding dy/dx = 0 to find the minimum. The reason I've shown no working, is that I don't know where to start exactly. If I could find a formula I'm sure I could do the rest, but I'm not sure where to go at the moment.
Please help

 

[tkhunny]

I'm not exactly sure, as I just don't know how to do it, but it probably concerns finding a formula to describe cost, then finding dy/dx = 0 to find the minimum. The reason I've shown no working, is that I don't know where to start exactly. If I could find a formula

This is what you are supposed to be learning. Of course you can solve if you had a formula given to you. That's why you've been studying Algebra. You should be able to create the formula. That's why you studied Geometry.

Give that drawing another look. Think about the Pythagorean Theorem. Right now, the cost is $9500*50 + $4700*120. Do you see that?

If you move point B around a little, you lose one Right Triangle, but gain another and can still build the cost relationship. For example, move B 40 km toward C. BC is now 80 km. What is AB? We still have A(OldB) = 50 km. We have B(OldB) = 40. Pythagoras tells us that \(\displaystyle AB^{2} = 50^{2} + 40^{2}\) so that AB = 64.03124237 km and we have another cost equation, $9500*60.03124237 + $4700*90. Is this more or less than the first try?

How can you generalize that example?

You MUST learn to think it through. You will NOT always be given a formula.

 

[galactus]

Choose an arbitrary point on the shore we'll call P.

The, the distance from P to A(the oil rig) is given by Pythagoras. Let the distance the from B to P be x

Then the remaining pipe along the shore will be 120-x=CP.

The total length of pipe will be CP+PA

\(\displaystyle PA=\sqrt{x^{2}+2500}\)

So, we have \(\displaystyle 9500\sqrt{x^{2}+2500}+4700(120-x)\)

This is what we must minimize.

 

[tkhunny]

Okay, naur, now that the formula hs been dropped in your lap it seems I have to find some other way to encourage you to think.

Are you SURE the point "p" doesn't go on the other side of point "B", away from Point "C".

 

[Naur]

tkhunny, it's not that I haven't tried or don't know how to do it. It's only that I've not seen a question like this before. All I need is to figure it out and do it a few times, and I'll get the hang of it, but it's the figuring out part that's the hardest.

galactus, I've had trouble finding the minimum of your formula. I differentiated to 9500x*(2500+x^2)^1/2 + (120-x)*4700, but need to find a value for x. Could you please point me in the right direction?

 

[galactus]

Sure, I'll help. I will do a little and then you finish?. If you get stuck let me know. Okey-doke?.

We have to differentiate our function:

\(\displaystyle C(x)=9500(x^{2}+2500)^{\frac{1}{2}}+4700(120-x)\)

Where are you having trouble with this?. It's just a little chain rule for the most part.

\(\displaystyle C'(x)=9500(\frac{1}{2}(x^{2}+2500)^{\frac{-1}{2}})(2x)-4700\)

\(\displaystyle =\frac{9500x}{\sqrt{x^{2}+2500}}-4700\)

Now, it's differentiated. Can you set to 0 and solve for x?.

 

[Naur]

Ooh, turns out I wasn't having trouble at all.
I got to that stage just fine before I posted, but I was having trouble finding a value for X. I had someone from school show me what I was doing wrong, and using the graphics calculator now I've done it.
My value for x is 28.46, and I can do the rest from there.

It's good too, because after I played around with that differentiated equation, I realized it had to be right, but couldn't get the right x value. Now that I've got it I'm very happy.
Thanks so much for your help !

 

[tkhunny]

tkhunny, it's not that I haven't tried or don't know how to do it.

Don't know how to do what?

It's only that I've not seen a question like this before.

That is exactly my point. You have been studying mathematics for some time if you are in calculus. You have the tools to solve problems that are new to you. What kind of social progress is there if we solve ONLY old problems. SOMEONE has to solve new problems. You would do well to adopt this philosophy.

All I need is to figure it out

You don't mean this. You want someone to show you. I would be delighted if you would "figure it out".

it's the figuring out part that's the hardest.

Exactly my point. That's the interesting part. Why let someone else do it for you?

 

[DR._Glockman]

Oil companies must be sold on "The Calculus" after seeing this problem.

 

[tkhunny]

I'm guessing the assumption of uniform cost in ANY terrain would be a hard sell!