BMAD Posted June 25, 2014 Report Share Posted June 25, 2014 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 Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted June 29, 2014 Report Share Posted June 29, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted June 25, 2014 Author Report Share Posted June 25, 2014 (edited) 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 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted June 28, 2014 Author Report Share Posted June 28, 2014 hint: pythagoras is a straightforward approach Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 28, 2014 Report Share Posted June 28, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 kukupai Posted June 28, 2014 Report Share Posted June 28, 2014 (edited) Edited June 28, 2014 by kukupai Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted June 28, 2014 Report Share Posted June 28, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted June 28, 2014 Report Share Posted June 28, 2014 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. Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 29, 2014 Report Share Posted June 29, 2014 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 Quote Link to comment Share on other sites More sharing options...

0 Yoruichi-san Posted June 29, 2014 Report Share Posted June 29, 2014 Actually you're right, the answer is 1.407, apparently my algebra with square roots sucks ;P Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted June 29, 2014 Report Share Posted June 29, 2014 Agree with Y-san's answer. Avoiding tortuous trig derivatives: Put P at (2, y). sin B + sin C = 1 (Condition for extremum is dy/dBP + dy/dCP = dy/dy) 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^{o}C = 25.12653497^{o} y = 2 tan B = 1.406997295 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted June 29, 2014 Author Report Share Posted June 29, 2014 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. 1 Quote Link to comment Share on other sites More sharing options...

0 harey Posted June 30, 2014 Report Share Posted June 30, 2014 (edited) 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 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted June 30, 2014 Author Report Share Posted June 30, 2014 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. Quote Link to comment Share on other sites More sharing options...

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## 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

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