minimum distance

[BMAD]

A point P needs to be located somewhere on the line AD so that the total length L of cables linking P to the points A, B, and C is minimized. What is the minimum value of L?

My image won't upload so you can see it here:

https://docs.google.com/file/d/0B0PFoZbqZhFCZUFicy10Q29od0U/edit?usp=docslist_api

[BMAD]

For those who can't access this image:

There is a triangle BPC where on one line segment BC there is a point D such that BD=2 and DC=3. Extending through P is a line segment going from some point A to D which is a distance of 5 and perpendicular to BC. Find the minimum total distance of all the "cables" by placing p in an optimal place.

Edited June 25, 2014 by BMAD

[BMAD]

hint: pythagoras is a straightforward approach

[plasmid]

Would anyone who's able to access the image be able to attach it as a figure?

From the description, I'm not sure what the cables are whose distance needs to be minimized.

[kukupai]

Edited June 28, 2014 by kukupai

[bonanova]

I got the same figure.

But, with plasmid, I don't know where the cables are.

Assumption (from post 1): We are to minimize PA + PB + PC.

[Yoruichi-san]

The point is placed ~0.099917m or about 10cm from the bottom. Too lazy to write up the equations, basically used calculus to find the minimum (can never get enough calculus ;P)

The height of A really doesn't matter, since that term disappears when you take the derivative, which makes sense, if you think about it.

[plasmid]

Thanks, kukupai.

I haven't fully worked out the equations, but I don't think that's it.

Using DP = 0.1 m, I get

AP = 4.9 m

BP > 2 m

CP > 3 m

Sum > 9.9 m

Using DP = 1 m, I get

AP = 4 m

BP = sqrt(5) m

CP = sqrt(10) m

Sum < 9.4 m

[Yoruichi-san]

Actually you're right, the answer is 1.407, apparently my algebra with square roots sucks ;P

[bonanova]

Agree with Y-san's answer.

Avoiding tortuous trig derivatives:

Put P at (2, y).

1. sin B + sin C = 1 (Condition for extremum is dy/dBP + dy/dCP = dy/dy)
2. tan B / tan C = 3/2 (Inspection)

Guess B reasonably, say 45^o.

Solve [1] for C. Check [2].

Increment B by small amount and repeat.

Update B based on change.

Repeat a few times.

B = 35.12633457^oC = 25.12653497^o

y = 2 tan B = 1.406997295

[BMAD]

I got the same figure.

But, with plasmid, I don't know where the cables are.

Assumption (from post 1): We are to minimize PA + PB + P

yes. PA + PB + PC = L

Actually you're right, the answer is 1.407, apparently my algebra with square roots sucks ;P

I am not sure what 1.407 is referring to but i get a different answer.

[Yoruichi-san]

The height of point P, i.e. the distance between P and D.

The total distance of rope or whatever was used is, as a function of the height of P, which is denoted x:

f(x)=sqrt(x^2+4)+sqrt(x^2+9)+(5-x)

Taking the derivative is pretty simple:

f'(x)=x/sqrt(x^2+4)+x/sqrt(x^2+9)-1

Set f'(x)=0 to find the minimum, and after a giant mess of algebra (which apparently I suck at :/) or plugging it into a numerical solver , you'll get a root of x=1.407.

[harey]

I get the same result as bonanova and Yoruichi-san.

Wanting to avoid algebraic errors, I asked Wolfram Alpha:

http://www.wolframalpha.com/input/?i=minimize+%28%28x*x%29%2B4%29**%281%2F2%29%2B%28%28x*x%29%2B9%29**%281%2F2%29%2B5-x&lk=4&num=5

Edited June 30, 2014 by harey

[BMAD]

The height of point P, i.e. the distance between P and D.

The total distance of rope or whatever was used is, as a function of the height of P, which is denoted x:

f(x)=sqrt(x^2+4)+sqrt(x^2+9)+(5-x)

Taking the derivative is pretty simple:

f'(x)=x/sqrt(x^2+4)+x/sqrt(x^2+9)-1

Set f'(x)=0 to find the minimum, and after a giant mess of algebra (which apparently I suck at :/) or plugging it into a numerical solver , you'll get a root of x=1.407.

I see the issue, I was looking for L the total minimal length of PA,PB,PC while you found the length of DP. Nice job.