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Applied Optimization

  • Thread starter ciubba
  • Start date
  • #1
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Homework Statement


An isosceles triangle has a base of length 4 and two sides of length 2sqrt(2). Let P be a point on the perpendicular bisector of the base. Find the location P that minimizes the sum of the distances between P and the three vertices.

Homework Equations


N/A

The Attempt at a Solution



Putting this on the cartesian coordinate system leaves me with one vertex, v1, at (0,0), v2 at (2,sqrt(2)) and v3 at (4,0).

The distance between the vertices and P would then be [tex]D_1=(2-x)^2+(\sqrt{2}-y)^2 \; \; D_2=(x-0)^2+(\sqrt{2}-y)^2 \; \; D_3=(4-x)^2+(0-y)^2[/tex] Their sum is my objective function, so [tex]D_t=(2-x)^2+(\sqrt{2}-y)^2 + (x-0)^2+(\sqrt{2}-y)^2 + (4-x)^2+(0-y)^2[/tex]

I'm assuming that I can come up with a constraint by similar triangles, but this seems like an incredibly obtuse way of solving this problem. Could someone point me to a better direction?
 

Answers and Replies

  • #2
57
7
Choose a coordinate system so that the perpendicular bisector becomes your y-axis. That would simplify things.
 
  • #3
65
2
Hmm, that's a good idea.

So, v1=(-2,0) , v2=(0,sqrt(2)) , v3=(2,0)

Edit: Still number crunching
 
Last edited:
  • #4
65
2
Okay, I get

[tex] D_1=4+y^2, \; D_2=(\sqrt{2}-y)^2, \; D_3=4+y^2[/tex]

Thus, the objective function, their sum, is [tex]D_t=(\sqrt{2}-y)^2+8+y^2[/tex]

[tex]D_t'=6 y-2 \sqrt{2}[/tex]

Which has a root at [tex]y=\frac{\sqrt{2}}{3}[/tex]

Unfortunately, that is the reciprocal of the book's answer. Where did I mess up?
 
  • #5
57
7
I think your coordinates for v2 are not correct. if the side length is 2sqrt(2), then v2 would be 2, right?
 
  • #6
57
7
Also, your values of D1, D2, and D3 are the square of the distances, so you have to take the square root.
 
  • #7
65
2
Doing that gives me the right answer-- thanks!
 
  • #8
57
7
Any time man! :D
 

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