BMAD 65 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 post Share on other sites

0 Solution Yoruichi-san 116 Posted June 29, 2014 Solution 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 post Share on other sites

0 BMAD 65 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 post Share on other sites

0 BMAD 65 Posted June 28, 2014 Author Report Share Posted June 28, 2014 hint: pythagoras is a straightforward approach Quote Link to post Share on other sites

0 plasmid 41 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 post Share on other sites

0 kukupai 3 Posted June 28, 2014 Report Share Posted June 28, 2014 (edited) Edited June 28, 2014 by kukupai Quote Link to post Share on other sites

0 bonanova 85 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 post Share on other sites

0 Yoruichi-san 116 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 post Share on other sites

0 plasmid 41 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 post Share on other sites

0 Yoruichi-san 116 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 post Share on other sites

0 bonanova 85 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 post Share on other sites

0 BMAD 65 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 post Share on other sites

0 harey 8 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 post Share on other sites

0 BMAD 65 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 post Share on other sites

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## BMAD 65

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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