Lagrange multiplication, find the point (x,y) of the plane in which the sum of the squares of the distances to the points (0,1), (0,0), (2,0) is minimal?
I don't understand if I'm supposed to use those 3 coordinates to generate an equation that is supposed to be my constraint?
To be honest, i don't understand what they mean by the squares of the distances.
On
The distance from point (x, y) to (0, 1) is $\sqrt{(x- 0)^2+ (y- 1)^2}$ so the "square of the distance" is $x^2+ (y-1)^2$. The distance from (x, y) to (0, 0) is $\sqrt{(x- 0)^2+ (y- 0)^2}$ so the "square of the distance" is $x^2+ y^2$. The distance from (x, y) to (2, 0) is $\sqrt{(x- 2)^2+ y^2}$ so the "square of the distance" is $(x- 2)^2+ y^2$. The "sum of the squares of the distances" is $x^2+ (y- 1)^2+ x^2+ y^2+ (x- 2)^2+ y^2= x^2+ y^2- 2y+ 1+ x^2+ y^2+ x^2- 4x+ 4+ y^2= 3x^2+ 3y^2- 2y- 4x+ 5$.
$f(x, y)= 3x^2+ 3y^2- 2y- 4x+ 5$ is the value you want to minimize. I am not clear what this has to do with "Lagrange multipliers" because you have not given any "constraints".
The partial derivatives are $f_x= 6x- 4= 0$ and $f_y= 6y- 2= 0$. The point $\left(\frac{2}{3}, \frac{1}{3}\right)$ satisfies those equations.
I'm sorry, but how can you know about "Lagrange multipliers", but not know what "the square of the distance" means? This is like discussing Shakespeare, but being flummoxed by "See Spot Run".
The distance between two points $(x, y)$ and $(a, b)$ is $\sqrt{(x-a)^2 + (y - b)^2}$. Obviously, the square of the distance is just $(x-a)^2 + (y - b)^2$.
Now for the point $(x,y)$, add up the square of the distance from $(x,y)$ to $(0,1)$, to $(0,0)$, and to $(2, 0)$. Then minimize that sum. You don't need Lagrange multipliers to do it. Just the set the two partial derivatives to $0$.