Five friends live on the same street. Their houses are at points A, B, C, D, and E, with the distances shown.
The five friends decide to meet at point P so that the total walking distance for all five friends is minimized. What is AP?
+128460 Here's my attempt at this one .....
Note
P = C ...... total distances walked = 9 + 4 + 2 + 5 = 20
P = midpoint of BC .....total distances walked = 7 + 2 + 2 + 4 + 7 = 22
P = one unit to the right of B.....total distances walked = 6 + 1 + 3 + 5 + 8 = 23
P = one unit to the left of C ....total distances walked = 8 + 3 + 1 + 3 + 6 = 21
P = one unit to the right of C .....total distances walked = 10 + 5 + 1 + 1 + 4 = 21
Let P be somewhere to the left of C
With a little help from WolframAlpha
Min
sqrt(9-x)2 + sqrt (4-x)^2 + sqrt (x^2) + sqrt (2 + x)^2 + sqrt (5 + x)^2 = 20 and x = 0 ....P = (9,0) , AP = 9
Now....let P be somewhere to the right of C
Min
sqrt (9 + x)^2 + sqrt (4 + x)^2 + sqrt (x^2) + sqrt (2 -x)^2 + sqrt (5 - x)^2 = 20 and x = 0....P =(9,0), AP = 9
So...it appears that the walking distances are minimized if all meet at C = P
