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Homework Help: Relative extrema

  1. Jul 3, 2003 #1
    Please help me?? I'm having great difficulty solving this question.

    Find all relative extrema of x^2y^2 subject to the constraint 4x^2+y^2=8. Do this in two ways:

    a)Use the constraint to eliminate a variable
    b)Use the method of Lagrange multipliers.

    Your help will be greatly appreciated.
     
  2. jcsd
  3. Jul 4, 2003 #2

    HallsofIvy

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    From the constraint y^2= 8- 4x^2 so we can rewrite the object function as f(x,y)= x^2y^2= x^2(8- 4x^2)= 8x^2- 4x^4= F(x).

    Now find F' and set it equal to 0.

    To use Lagrange Multipliers, rewrite the object function
    as F(x,y)= x^2y^2+ lambda(4x^2+ y^2- 8) and take the gradient:

    grad F= (2xy^2+ 4 lambda x)i+ (2x^2y+ 2 lamda y)j

    Set that equal to 0i+ 0j and with 4x^2+ y^= 8 you have three equation with which to determine x,y, and lambda.
     
  4. Jul 9, 2003 #3
    Hi HallsofIvy, since you last posted your reply to my problem I've been working on finding the solution.

    I managed to find all the relative extrema for a) by eliminating a variable.

    But I'm having real trouble finding all the relative extrema for b) using the method of lagrange multipliers?

    From your advice, I set the grad F= (2xy^2+ 4 lambda x)i+ (2x^2y+ 2 lamda y)j equal to zero and got

    2xy^2+ 4 lambda x=0
    2x^2y+ 2 lambda y=0
    4x^2+ y^2= 8

    solving for x,y and lambda:
    y^2=8-4x^2 so y=sqrt(8-4x^2)
    4x^2=8-y^2 so x^2=(8-y^2)/4 and x=sqrt[(8-y^2)/4]

    so substituting y^2, x^2, x, and y into 2x^2y+ lambda x=0 I get;
    2x(8-4x^2) + 4 lambda x =0

    This is where it all gets a little bit weird!!!!

    Can anyone show me how they would approach this and help solve my dilemma?

    Your help is much appreciated.
    Regards
     
  5. Jul 9, 2003 #4

    HallsofIvy

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    I was about to chastise you for doing the derivative wrong when I realized you had copied MY mistake. The derivative of 4x^2 is, of course, 8x not 4x so 2xy^2+ 4 lambda x=0 should be 2xy^2+ 8lambda x= 0.
    the three equations are
    2xy^2+ 8 lambda x=0
    2x^2y+ 2 lambda y=0
    4x^2+ y^2= 8

    The way I would solve them is to rewrite the first equation as 2x(y^2+ 2lambda)= 0. Either x= 0 or y^2+ 2lambda= 0.

    If x= 0 then y^2= 8 so y= +/- 2sqrt(2) (from the last equation).

    The middle equation can be rewritten (4x^2+ 2lambda)y= 0. Either y= 0 or 4x^2+ 2lambda= 0.

    If y= 0 then 4x^2= 8 so x= +/- sqrt(2) (from the last equation).

    If neither x nor y is zero then we have y^2+ 2lambda= 0 so
    lambda= -(1/2)y^2 and 4x^2+ 2lambda= 0 so lambda= -2 x^2. Those together give -2x^2= -(1/2)y^ or y^2= 8x^2.

    Putting that into 4x^2+ y^2= 8, 4x^2+ 8x^2= 8 or 12x^2= 8.

    x^2= 8/12= 2/3 so x= +/- sqrt(2/3) and 4(2/3)+ y^2= 8 so
    y^2= 8- 8/3= 16/3 and y= +/- 4sqrt(1/3).

    The "critical points" are (sqrt(2), 0), (sqrt(2), 0),
    (0,2 sqrt(2)), (0, -2 sqrt(2)), (sqrt(2/3), 4 sqrt(1/3)),
    (-sqrt(2/3), 4 sqrt(1/3)), (sqrt(2/3), -4 sqrt(1/3)), and
    (-sqrt(2/3), -4 sqrt(1/3)).
     
  6. Jul 9, 2003 #5

    HallsofIvy

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    BLAST IT! I accidently hit the "post" button while I was in the middle of correting my mistake.

    I was about to chastise you for doing the derivative wrong when I realized you had copied MY mistake. The derivative of 4x^2 is, of course, 8x not 4x so 2xy^2+ 4 lambda x=0 should be 2xy^2+ 8lambda x= 0.
    the three equations are
    2xy^2+ 8 lambda x=0
    2x^2y+ 2 lambda y=0
    4x^2+ y^2= 8

    The way I would solve them is to rewrite the first equation as
    2x(y^2+ 4lambda)= 0. Either x= 0 or y^2+ 4lambda= 0.

    If x= 0 then y^2= 8 so y= +/- 2sqrt(2) (from the last equation).

    The middle equation can be rewritten 2(x^2+ lambda)y= 0. Either y= 0 or x^2+ lambda= 0.

    If y= 0 then 4x^2= 8 so x= +/- sqrt(2) (from the last equation).

    If neither x nor y is zero then we have y^2+ 4lambda= 0 so
    lambda= -(1/4)y^2 and x^2+ lambda= 0 so lambda= -x^2. Those together give -x^2= -(1/4)y^ or y^2= 4x^2.

    Putting that into 4x^2+ y^2= 8, 4x^2+ 4x^2= 8 or 8x^2= 8 and x= +/1.

    If x= +/- 1, then 4+ y^2= 8 or y^2= 4 so y= +/- 2.

    The "critical points" are (sqrt(2), 0), (sqrt(2), 0),
    (0,2 sqrt(2)), (0, -2 sqrt(2)), (1, 2), (-1, 2), (1, -2), and (-1, -2)
    , exactly the same result you get by eliminating a variable.
     
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