# Applied Optimization

## 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.

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 $$D_1=(2-x)^2+(\sqrt{2}-y)^2 \; \; D_2=(x-0)^2+(\sqrt{2}-y)^2 \; \; D_3=(4-x)^2+(0-y)^2$$ Their sum is my objective function, so $$D_t=(2-x)^2+(\sqrt{2}-y)^2 + (x-0)^2+(\sqrt{2}-y)^2 + (4-x)^2+(0-y)^2$$

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?

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Choose a coordinate system so that the perpendicular bisector becomes your y-axis. That would simplify things.

Hmm, that's a good idea.

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

Edit: Still number crunching

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Okay, I get

$$D_1=4+y^2, \; D_2=(\sqrt{2}-y)^2, \; D_3=4+y^2$$

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

$$D_t'=6 y-2 \sqrt{2}$$

Which has a root at $$y=\frac{\sqrt{2}}{3}$$

Unfortunately, that is the reciprocal of the book's answer. Where did I mess up?

I think your coordinates for v2 are not correct. if the side length is 2sqrt(2), then v2 would be 2, right?

Also, your values of D1, D2, and D3 are the square of the distances, so you have to take the square root.

Doing that gives me the right answer-- thanks!

Any time man! :D