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Homework Help: Lagrange multiplier question

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the points on the level surface xy2z4=1 that are closest to the origin.

    2. Relevant equations

    Lagrange's method for finding extrema

    3. The attempt at a solution

    If I have a level surface F(x,y,z)=c, it's points closest to the origin will be the ones in which the gradient vector points to the origin. A generic vector pointing to/from the origin is G=<x,y,z>, so F must be a scalar multiple of G.

    I come up with a system of equations

    ßx=y2z4
    ßy=2xyz4
    ßz=4x2z3
    xy2z4=1.

    I can first simplify a little bit.

    ßx=y2z4
    ß=2xz4
    ß=4x2z2


    I can set the 2nd and 3rd equations equal.

    2xz4=4x2z2
    ----> x= z2/2

    I can plug that x into the first 2 equations.

    (y2z4)/[z2/2]=2[z2/2]z4
    ----> y = +/- √(z4/2)

    Plugging those into the constraint xy2z4=1

    ----> z=4(1/10).

    Am I right? What is the most straight-forward way of solving such a problem?
     
  2. jcsd
  3. May 4, 2010 #2

    lanedance

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    looking at [itex] xy^2z^4=1 [/itex] can make the following observations that may help later
    [itex] x =\frac{1}{y^2z^4} [/itex]
    so x > 1 and [itex] x,y,z \neq 0[/itex]
     
    Last edited: May 5, 2010
  4. May 4, 2010 #3

    lanedance

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    then i think you're doing the correct thing with the lagrange multipliers

    the gradient point to the origin comes from minimising [itex] g(x,y,z) = x^2 + y^2 + z ^2 [/itex] subject to the constraint [itex] f(x,y,z) = xy^2z^4 = 1[/itex]

    then taking the gradient
    [tex] \nabla g(x,y,z) = \nabla(x^2 + y^2 + z ^2) = 2(x,y,z) [/tex]
    note is teh direction from the origin to the point as you say

    [tex] \nablaf(x,y,z) = \nabla(xy^2z^4) = (y^2z^4 ,2xyz^4 ,4xy^2z^3 )[/tex]

    so i think your 3rd equation is missing a y^2
     
  5. May 5, 2010 #4

    HallsofIvy

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    Since the value of the lagrange multiplier, [itex]\beta[/itex], is not necessary to the solution, I often find it simplest to start by dividing equations to eliminate the multiplier.

    Dividing [itex]\beta x= y^2z^4[/itex] by [itex]\beta y= 2xyz^4[/itex] gives
    [tex]\frac{x}{y}= \frac{y^2z^4}{2xyz^4}= \frac{y}{2x}[/tex]
    so that [itex]2x^2= y^2[/itex] and [itex]y= \sqrt{2}x[/itex].

    Dividing [itex]\beta y= 2xyz^4[/itex] by [itex]\beta z= 4x^2z^3[/itex] gives
    [tex]\frac{y}{z}= \frac{2xyz^4}{4x^2z^3}= \frac{yz}{2x}[/tex]
    so that [itex]2x= z^2[/itex] and [itex]x= z^2/2[/itex] and then [itex]y= \sqrt{2}{x}= z^2/\sqrt{z}[/itex].

    Putting those into the equation of the surface, [itex]xy^2z^4= 1[/itex] gives [itex](z^2/2)(z^4/2)(z^4)= z^10/4= 1[/itex] so that [itex]z= (1/4)^{1/10}[/itex].

    Check my arithmetic.
     
  6. May 5, 2010 #5

    lanedance

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    i don't think equation 3 is correct - shouldn't the equation set be
    [tex]\beta x= y^2z^4[/tex]
    [tex]\beta y = 2xyz^4 [/tex]
    [tex]\beta z= 4xy^2z^3 [/tex]
    [tex]1= xy^2z^4 [/tex]
     
  7. May 5, 2010 #6

    lanedance

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    also as the surface is symmetric about the xy and xz axes, i'd think there should be at least 2 points... if not 4...
     
    Last edited: May 5, 2010
  8. May 5, 2010 #7

    lanedance

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    1 div 2
    [itex]2x^2= y^2[/itex]

    1 div 3
    [tex]\frac{x}{z}= \frac{y^2z^4}{4xy^2z^3 }= \frac{z}{4x}[/tex]
    [tex]4x^2 = z^2[/tex]
     
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