Explore BrainMass

Explore BrainMass

    Lagrangian multipliers

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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 March 4, 2021, 6:03 pm ad1c9bdddf
    https://brainmass.com/math/basic-algebra/lagrangian-multipliers-25918

    Solution Preview

    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 ...

    Solution Summary

    This shows how to use Lagrangian multipliers to find the points on the ellipse with the given characteristics.

    $2.49

    ADVERTISEMENT