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Homework Help: Extreme values (Lagrange multipliers)

  1. Mar 18, 2010 #1
    determine, if any, the maximum and minimum values of the scalar field f (x, y) = xy subject to the constraint [tex]4x^2{}[/tex]+[tex]9y^2{}[/tex]=36

    The attempt at a solution

    using Lagrange multipliers, we solve the equations [tex]\nabla[/tex]f=[tex]\lambda[/tex][tex]\nabla[/tex]g ,which can be written as




    or as




    it's pretty much all done but can somebody solve this? cause i have some doubts about which are the extreme points
  2. jcsd
  3. Mar 18, 2010 #2
    I would say your [tex]g(x,y) = 36[/tex] is a little suspicious. If this were the case then the two statements above would be false since [tex]g_x[/tex] and [tex]g_y[/tex] are 0.
  4. Mar 18, 2010 #3
    no, in fact g(x,y)=[tex]4x^2{}[/tex]+[tex]9y^2{}[/tex]=36 but well, I wrote it that way because is the way my book does
  5. Mar 18, 2010 #4
    We already know that (0,0) is a turning point which has a value f (0,0) = 0.

    Leaving that aside for now, solving the set of equations that you listed:
    [tex]y = 8\lambda x[/tex]
    [tex]x = 18\lambda y[/tex]
    [tex]4x^2 + 9y^2 = 36[/tex]
    would yield one or more points.

    Then find the value of f(x,y) at each of these points and compare.
  6. Mar 18, 2010 #5
    Last edited by a moderator: May 4, 2017
  7. Mar 19, 2010 #6
    Firstly, careless mistake: y is [tex]+/- \frac{2}{3}\sqrt{\frac{18}{5}}[/tex] as you've written on the third line, but in carrying it over into f(x,y), you mixed up the denominator and numerator.

    Secondly, [tex]\lambda = +/- \frac{1}{12}[/tex], so there exist two other solutions:
    [tex](x,y) = (\sqrt{\frac{18}{5}}, - \frac{2}{3}\sqrt{\frac{18}{5}})[/tex]
    [tex](x,y) = (- \sqrt{\frac{18}{5}}, \frac{2}{3}\sqrt{\frac{18}{5}})[/tex]
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