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 Multipliers to find the Maximum and Minimum values

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Use Lagrange Multipliers to find the Maximum and Minimum values of f(x,y) = x2-y.
    Subject to the restraint g(x,y) = x2+y2=25

    2. Relevant equations

    gradient f(x,y)= gradient g(x,y)

    3. The attempt at a solution

    I have found the gradients of f and g to be

    f(x,y) = 2xi + -1j
    g(x,y) = 2xi + 2yj


    I have put these to gether with the constraint to find the simutaneous equation

    2x = lambda 2x
    -1 = lambda 2y
    x2+y2 = 25

    Have I got this system right so far ?

    I'm a bit concerned about the 2x = lambda 2x part.
    I'm not sure if this is right or I should have used 2x = lambda 2y ?

    In the original system I have Lambda = 1 and y = -1/2

    Any help greatly appreciated
    Brendan
     
  2. jcsd
  3. Apr 4, 2009 #2
    Everything looks right to me, and no, you definitely should not have used [itex]2x=\lambda 2y[/itex]. The general system for a function [itex]f(x,y)[/itex] constrained to [itex]g(x,y) = k[/itex] is:
    [tex] \begin{align*}
    f_x &= \lambda g_x \\
    f_y &= \lambda g_y \\
    g &= k
    \end{align*}[/tex]
    So you have done this exactly right. You have found [itex]\lambda=1,y=-1/2[/itex] correctly. What about x?
     
    Last edited: Apr 4, 2009
  4. Apr 4, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What are you concerned about? I think you are doing fine. 2x = lambda 2y isn't right at all. Why would you think it is?
     
  5. Apr 4, 2009 #4
    g(x,y) = x2+y2=25

    Now we have y = -1/2

    so x = sqrt(y2+25) or -sqrt(y2+25)

    = sqrt(101)/2 or -sqrt(101)


    So we have two solutions
    f(x,y)= x2-y = f(sqrt(101)/2,-1/2) = 103/4 = 25.75
    f(x,y)= x2-y = f(-sqrt(101)/2,-1/2) = 103/4 = 25.75

    So they are both the maximum
     
  6. Apr 4, 2009 #5
    Whoops, you have a sign error. It should have been [itex]x = \pm\sqrt{25-y^2}[/itex].
     
  7. Apr 5, 2009 #6
    g(x,y) = x2+y2=25

    Now we have y = -1/2

    so x = sqrt(25-y2) or -sqrt(25-y2)

    = sqrt(99/4) or -sqrt(99/4)


    So we have two solutions
    f(x,y)= x2-y = fsqrt(25-y2) ,-1/2) = 101/4= 25.25 (maximum)
    f(x,y)= x2-y = f(-sqrt(25-y2) ,-1/2) = -97/4= -24.25 (Minimum)

    Is that better?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook