Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Lagrange Multipliers

  1. Dec 6, 2005 #1
    Hi, I'm having trouble with the following question.

    Q. Find the maximum and minimum of the function f(x,y) = x^2 + xy + y^2 on the circle x^2 + y^2 = 1.

    I started off by writing:

    Let g(x,y) = x^2 + y^2 then [tex]\nabla f = \lambda \nabla g,g\left( {x,y} \right) = 1[/tex]

    \Rightarrow 2x + y = 2\lambda x...(1)

    2y + x = 2\lambda y...(2)

    x^2 + y^2 = 2...(3)

    I'm not sure how to solve this system of equations. I've got the impression that generally an explicit value for lamda is not needed. I'd normally start off by considering the possible cases.

    Here, if x = 0 in (1) then from (2) I get y = 0 so that (x,y) = (0,0). However this contradicts (3) and it isn't on the circle so I ignore this point. But now I'm stuck. Which values of x or y should I try now? Any help would be good thanks.
    Last edited: Dec 6, 2005
  2. jcsd
  3. Dec 6, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Add equations (1) and (2).

    Is your constraint g(x,y)=1 or g(x,y)=2?
  4. Dec 6, 2005 #3


    User Avatar
    Science Advisor

    A fairly common way of handling equations like that, where you don't need to find [itex]\lambda[/itex] is to divide one equation by the other.
    You get [itex]\frac{2x+y}{2y+x}= \frac{x}{y}[/itex] or y(2x+y)= x(2y+ x). That obviously gives x2= y2 so that either y= x or y= -x and that, together with x2+ y2= 1 gives you the answer.

    By the way, since we have the constraint x2+ y2= 1, f(x,y)= x2+ xy+ y2 immediately reduces to
    f(x,y)= 1+ xy on that circle. That makes the calculations a little easier.
  5. Dec 6, 2005 #4
    Thanks for the help Galileo and HallsofIvy.

    The constraint is g(x,y) = 1, I'll fix it up now.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook