1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrange multpliers

  1. Oct 15, 2011 #1
    Use lagrange multipliers to find max/min of
    f(x,y)=x^2+6y
    subject to
    x^2-y^2=5

    grad f =λgrad g
    2x=2xλ, λ=1
    6=-2yλ, λ=-3/y
    1=-3/y, y=-3

    x^2-(-3)^2=x^2-9=5
    x^2=14
    x=+/-√14
    two points are √14, -3 and -√14, -3
    plugging both points into f(x,y) gives me the same answer. now what?
     
  2. jcsd
  3. Oct 15, 2011 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper


    There are two constrained minima but no constrained maxima. Why not? Well, we can find values of x and y that together go to infinity along the curve g = 0, and when we do that we get f --> +infinity.

    RGV
     
  4. Oct 15, 2011 #3
    thank you very much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lagrange multpliers
  1. Lagrange with integral (Replies: 2)

  2. Lagrange Multipliers (Replies: 8)

  3. Lagrange Multipliers (Replies: 8)

  4. Lagrange's Identity (Replies: 7)

Loading...