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!

Homework Help: Maxima/Minima of extrema function f(x,y)=x

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the extrema of the function subject to the given constraint.

    f (x, y) = x

    x2 + 2y2 = 3


    2. Relevant equations
    det = fxx * fyy - (fxy)2
    If det > 0,
    fxx < 0 [tex]\Rightarrow[/tex] MAXIMUM
    fxx > 0 [tex]\Rightarrow[/tex] MINIMUM

    If det < 0,
    [tex]\Rightarrow[/tex] SADDLE POINT

    If det = 0m
    Inconclusive

    3. The attempt at a solution

    I appear to be struggling with this maxima/minima stuff. Previous questions were just correct. I had to look at the book's answers to figure a way there. I seem to be missing something...but here is my attempt:

    Since x2 + 2y2 = 3,

    x = [tex]\pm[/tex][tex]\sqrt{3 - 2y2}[/tex]

    Therefore,

    f(x, y) = [tex]\pm[/tex][tex]\sqrt{3 - 2y2}[/tex]

    At critical point,

    [tex]\nabla[/tex] f = (0, - [tex]\frac{2y}{\sqrt{3 - 2y2}}[/tex] )

    = (0, 0)


    Therefore, equating gives

    0 = 0

    &

    - [tex]\frac{2y}{\sqrt{3 - 2y2}}[/tex] = 0

    So extrema points at ([tex]\sqrt{3}[/tex], 0) & (-[tex]\sqrt{3}[/tex], 0) since y = 0.


    To find max/min/saddle, find det.

    fxx = 0
    fxy = 0

    det = fxx * fyy - (fxy)2
    det = 0 * fyy - (0)2
    det = 0


    Which is inconclusive. The book has ([tex]\sqrt{3}[/tex], 0) as a maximum and (-[tex]\sqrt{3}[/tex], 0) as a minimum.

    I don't know what I'm missing out here...

    Thanks in advance.
     
  2. jcsd
  3. Mar 22, 2010 #2

    Mark44

    Staff: Mentor

    Tip: Don't mix [ sup] or [ sub] tags inside [ tex] tags. Instead use ^{} for exponents and _{} for subscripts.
    I don't see how the equation above is helpful at all.
    What I think you are missing is some understanding of what f(x, y) = x looks like, which is a plane. Since a plane is a flat surface with no curvature, there are no dips or bumps. Also, since the domain of this plane is restricted to {(x, y) | x2 + 2y2 = 3}, any maximum or minimum points are attained at points the farthest away from the origin.
     
  4. Mar 22, 2010 #3
    Ohh...I see... I feel so silly ><

    Thanks a lot for that.



    If I may, I have another question regarding maxima and minima...

    The problem statement, all variables and given/known data

    Find local minima, maxima and saddle points of the function:

    f(x, y) = (x2 + 3y2) e1-x2-y2


    I started off finding the partial derivatives of x and y respectively and got:

    fx = 2xe1-x2-y2 ( 1 - x2 - 3y2 )
    fy = 2ye1-x2-y2 (3 - x2 - 3y2 )


    At critical points, (fx, fy) = (0, 0)

    Equating gives,

    2xe1-x2-y2 ( 1 - x2 - 3y2 ) = 0 --- (1)
    2ye1-x2-y2 (3 - x2 - 3y2 )= 0 --- (2)



    After that I'm kinda lost...I slithered my way to get (0, 0) but very unconvincingly. I would rather have some solid reasoning as to why.

    Like I mentioned earlier, I think I'm missing some understanding and maybe that's why I'm not getting this.

    The answers give
    Min at (0, 0),
    max at (0, 1), (0, -1),
    saddle at (-1, 0), (1, 0)

    Thanks!
     
  5. Mar 22, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    When you have a product equal to 0, then one of the factors must be equal to 0. For instance, from this equation

    [tex]2xe^{1-x^2-y^2}(1-x^2-3y^2) = 0[/tex]

    you can conclude that x=0 or 1-x2-3y2=0. Similarly, the other equation gives you y=0 or 3-x2-3y2=0. Now you just have to find the solutions for all four possible combinations:

    (1) x=0, y=0;
    (2) x=0, 3-x2-3y2=0;
    (3) y=0, 1-x2-3y2=0; and
    (4) 1-x2-3y2=0, 3-x2-3y2=0.
     
  6. Mar 22, 2010 #5
    Ooo, I see now. I think I managed to confuse myself by using how for eq (1),

    since x = 0,

    1-x2-3y2=0
    1 - 3y2 = 0

    which is definitely not looking good in terms of the answers. But I understand now =)

    One last thing, so for the 4th solution: (4) 1-x2-3y2=0, 3-x2-3y2=0.

    There isn't a solution for this yeah? Since all the variables x/y cancel out.
     
  7. Mar 22, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Right idea, but wrong equation. That's the one that has to hold if y is 0 and x isn't 0.
    Yup, that combination has no solutions.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook