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Lagrange Multipliers HELP PLEASE

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the maximum and minimum values of f = (x-1)^2 + (y-1)^2 on the boundary of the circle g = x^2 + y^2 = 45.


    2. Relevant equations

    f=(x-1)^2 + (y-1)^2
    g=x^2+y^2=45
    gradf(x,y)=lambda*gradg(x,y)

    3. The attempt at a solution

    gradf(x,y)=<2x-2,2y-4>
    gradg(x,y)=<2x,2y>
    (1) 2x-2=lambda*2x
    (2) 2y-4=lambda*2y
    (3)g=x^2+y^2=45

    solving the system for critical points:
    x = 1/(1-lambda) plug into g?
    y = 2/(1-lambda) plug into g?
    gives lambda = 1 - (1/x) and lambda = 1 - (2/y)
    set lambda = lambda:
    1 - (1/x) = 1 - (2/y)
    (1/x) = (2/y)
    y = 2*x
    here's where I get lost, how do i plus this back in? I shouldn't have to know the sqrt(45) to solve this. HELP
     
  2. jcsd
  3. Nov 29, 2008 #2

    gabbagabbahey

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    Well, IF [itex]y=2x[/itex], then [tex]x^2+y^2=x^2+(2x)^2=5x^2=45 \Rightarrow x= \pm 3[/tex]

    But(!) you've made a mistake earlier in your calculations: Is [itex]\vec{\nabla}f(x,y)[/itex] really [itex]\left< 2x-2, \, 2y-4 \right>[/itex]? :wink:
     
  4. Nov 29, 2008 #3
    whoops f(x,y) = (x-1)^2+(y-2)^2
    sorry about the typo
     
  5. Nov 29, 2008 #4
    once i've solved to find that x=(plus or minus)3, where do I plug it back in to find the critical points? do I plug it into (1) and solve for lamba? or do i plus it into (3) again and solve for y?
     
  6. Nov 29, 2008 #5

    gabbagabbahey

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    well, you plug it into y=2x first, so that you can find the corresponding y values....then you plug your (x,y) pairs into your expression for f and find out which ones are maximums and which are minumums.....for example, if I found that f(3,6)=2 and f(-3,-6)=4000 then I would know that the point (-3,-6) is a maximum and the point (3,6) was a minimum and that the maximum and minimum values of f were 4000 and 2 respectively.
     
  7. Nov 29, 2008 #6
    do i need to find any more critical points? or will that suffice.. thank you for your help, by the way. your explanations have been clear, concise and beyond enlightening.
     
  8. Nov 29, 2008 #7

    gabbagabbahey

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    Since all you are looking for are maximums and minimums, there are no other critical points to consider.
     
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