Use Lagrangian multipliers to find the points on the ellipse with the equation x^2+xy+y^2=3 that are closest to and farthest from the origin© BrainMass Inc. brainmass.com December 24, 2021, 5:04 pm ad1c9bdddf
Well, we consider a point (x, y) on the ellipse. Its distance to the origion would be x^2+y^2. Therefore, we want to optimize x^2+y^2 with respect to the constraint: x^2+ xy+ y^2= 3 or x^2+ xy+ y^2- 3= 0 which is the equation of the given ellipse. Let's define p to be the lagrange multiplier, then we will have:
F= x^2+ y^2+ p(x^2+ xy+ y^2- 3)
Now we take the derivative of F with respect to x, ya nd p respectively and put them equal to zero to find the critical points.
dF/dx= 2x+ 2px+ py= 0 (1)
dF/dy= 2y+ px+ 2py= 0 (2)
df/dp= x^2+ xy+ y^2- 3= 0 (3)
Ok, we have 3 equations. The first two equations are linear in x and y and ...
This shows how to use Lagrangian multipliers to find the points on the ellipse with the given characteristics.