Optimization - find point by minimising squared distance

[inner08]

Find the point in the plane 3x+2y+z=1 that is the closest to the origin by minimising squared distance. (I hope I translated this ok..)

I was thinking I would need to isolate a variable in the equation for the plane above then substitute it into the distance formula then do a partial derivative.

Something like... D=(x^2+y^2+z^2)^(1/2). Since it is squared I could just have (x^2+y^2+z^2).

z=1-2y-3x

D = x^2+y^2+(1-2y-3x)^2

= x^2+y^2+1-4y-6x+12xy+9x^2
=10x^2+y^2+1-4y-6x+12xy

f_x = 20x-6+12y = 0
y = (6-20x)/12


f_y = 2y-4+12x
y = (4-12x)/2

etc...x=1/4, y=1/2, z=-3/4

It seems ok because it works in the given equation(3x+2y+z=1) but I've never done a problem like this so if its wrong, I do hope I'm atleast on the right track. Hope someone can't let me know.

Thanks,

 

[mathman]

You made an error in your calculation of D. In squaring the expression for z, you omitted 4y^2. Therefore D should have 5y^2 not y^2. You should be able to get the right answer after that. It'll be (3,2,1)/14.

 

[tacman]

Your approach is correct. To minimize the squared distance, we can use the method of Lagrange multipliers. We need to minimize the function f(x,y,z) = x^2 + y^2 + z^2 subject to the constraint g(x,y,z) = 3x+2y+z-1 = 0. This can be written as the Lagrangian L = f(x,y,z) - λg(x,y,z) where λ is the Lagrange multiplier.

Taking partial derivatives and setting them equal to 0, we get:

f_x = 2x - 3λ = 0
f_y = 2y - 2λ = 0
f_z = 2z - λ = 0
g_x = 3
g_y = 2
g_z = 1

Solving these equations, we get x = 1/4, y = 1/2, z = -3/4. Substituting these values into the constraint, we get the point (1/4, 1/2, -3/4) as the closest point to the origin on the plane 3x+2y+z=1.

Your approach of using the distance formula and taking partial derivatives is also valid and leads to the same solution. Good job on solving this problem!