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Lagrange multipliers elliptic paraboloid

  1. Jul 3, 2003 #1
    Hi, I'm really stuck on this problem and I need some help??

    Here's the question:

    The intersection of the elliptic paraboloid z=x^2+4y^2 and the right circular cylinder x^2+y^2=1. Use Lagrange multipliers to find the highest and lowest points on the curve of intersection.

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

    HallsofIvy

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    By "highest and lowest points", I take it you mean maximize and minimze the z coordinate?

    In that case your "object function" is z and you have the conditions that z=x^2+4y^2 (or x^2+ y^2- z= 0) and x^2+y^2=1.

    Okay, look at F(x,y,z)= z+ lambda(x^2+ 4y^2- z)+ mu(x^2+y^2-1)
    (lambda and mu are the Lagrange multipliers).

    The gradient of F is
    grad F= (2 lambda x+ 2 mu x)i+ (8 lamda y+ 2 mu y)j+ (1- lambda)k

    So we have the 5 equations
    2 lambda x+ 2 mu x= 0
    8 lambda y+ 2 mu y= 0
    1- lambda= 0 (which kinda tells you what lamda is!)
    x^2+ 4y^2- z= 0 and
    x^2+ y^2= 1
    to solve for the 5 unknowns x, y, z, lambda, mu.

    I get that the maxima z occur at (-1,0,1), (1,0,1), (0, -1,1)
    and (0,l,1).

    Are you required to use Lagrange multipliers?

    It simpler to rewrite z= x^2+ 4y^2= x^2+y^2+ 3y^2= 3y^2+ 1 and maximize that and to rewrite z= x^2+ 4(1-x^2)= -3x^2+ 4 and minimize that.
     
  4. Jul 5, 2003 #3
    Hi HallsofIvy, thanks for helping me with my problem. I have a couple of questions I would like to ask you about it.

    1) Could you please explain step by step how you solved the 5 equations?
    2 lambda x+ 2 mu x= 0
    8 lambda y+ 2 mu y= 0
    1- lambda= 0 (which kinda tells you what lamda is!)
    x^2+ 4y^2- z= 0 and
    x^2+ y^2= 1
    to solve for the 5 unknowns x, y, z, lambda, mu.

    As I found it difficult to decipher what the values of x, y, z, lamda, and mu were?

    2) You got that the maxima z occur at (-1,0,1), (1,0,1), (0, -1,1) and (0,l,1).

    I'm assuming that (0,l,1) is a typo and is meant to be (0,1,1).

    As you have only found the maxima points, the above solution...
    How would you use lagrange to find the lowest (minima) points on the curve of intersection?

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

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Actually, I miswrote when I said "maximum". Obviously, y= 0 is a minimum of the function z= 1+ 3y^2.

    By letting y^2= 1- x^2, in x^2+ 4y^2= z, we get
    z= x^2+ 4(1- x^2)= 4- 3x^2. Then z'= -6x so x= 0 is a maximum for that function. If x= 0, then, from x^2+ y^2= 1, y= 1 or -1 and
    z= 0^2+ 4(1)= 4 is the maximum.


    Now, solving the equations from the Lagrange multiplier:

    From "lambda- 1= 0" we get lambda= 1.

    Then "2 lambda x+ 2 mu x= 0" become 2x+ 2mu x= 2x(1+ mu)= 0 so either x= 0 or mu= -1.

    If x= 0 then the equation x^2+ y^2= 1 becomes y^2= 1 so y= 1 or -1.
    If x= 0 and y= 1 or -1 then x^2+ 4y^2- z= 0 gives z= 4.

    If x is not 0, then mu= -1 so 8 lambda y+ 2 mu y= 0 becomes
    8y+ -2y= 0 or y= 0. With y= 0, x^2+ y^2= 1 gives x^2= 1 and x= 1 or x= -1. With x= 1 or -1, y= 0, z= 1.

    That is, (-1,0,1) and (1,0,1) give the MINIMUM values of z while

    (0,1,4) and (0,-1,4) give the maximum values of z.
     
  6. Jul 5, 2003 #5
    Hey cheers for your help HallsofIvy, I appreciate it.
     
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