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Lagrange Multipliers. All variables cancel

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data
    A cannonball is heated with with temperature distribution T(x,y,z)=60(y2+z2-x2).

    The cannonball is a sphere of 1 ft with it's center at the origin

    a) Where are the max and min temperatures in the cannonball, and where do they occur?


    2. Relevant equations
    [itex]\nabla[/itex]f=λ[itex]\nabla[/itex]g

    Where g is the the constraint and λ is the common ratio.

    [itex]\nabla[/itex]= fx i + fy j + fz k



    3. The attempt at a solution

    The cannonball is the restraint so
    λ[itex]\nabla[/itex]f = 2xλ i + 2yλ j + 2zλ k
    [itex]\nabla[/itex]T = -120x i + 120y j + 120z k

    2xλ = -120x, λ= -60
    2yλ = 120y, λ = 60
    2zλ = 120z, λ = 60

    I don't know where to go from here. All the variables canceled so I can't relate them to each other, and the λs are different.
     
    Last edited: Nov 29, 2011
  2. jcsd
  3. Nov 29, 2011 #2

    Dick

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    2*x*lambda=(-120*x) doesn't mean lambda=(-60). It means either x=0 OR lambda=(-60). You have to consider both cases.
     
  4. Nov 29, 2011 #3
    But couldn't x be any value? For x ≠ 0, λ=-60. And then when x = 0 we get 0=0 with no λ.

    If you do that for all the values you can get x=y=z=0, plug into the constraint and get the center of the circle, with no value for λ. How would that help find the max and the min?
     
  5. Nov 29, 2011 #4

    Dick

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    x=y=z=0 is an important point. You should keep it in mind. It's an interior critical point. But yes, if x is nonzero then lambda=(-60). What does that tell you about y and z? If x is zero then lambda could be 60. Still doesn't tell you anything about y and z. But now you have to use the constraint y^2+z^2-x^2=1. You have to think this all through. You don't necessarily get a finite number of extrema on the boundary. There might be a lot of them.
     
  6. Nov 29, 2011 #5
    Well if x=0 then the constraint becomes y^2+z^2=1, with λ=60. So would the maxima occur anywhere in the yz unit circle? Which would yield (0, y, ±√(1-y^2)) as the maxima?

    And then the smallest λ would occur at x≠0, and y=z=0? Which would give Minima on
    x: [-1,0) U (0,1]
    y=0
    z=0

    Or am I off-base here?
     
  7. Nov 29, 2011 #6

    Dick

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    Getting closer. So the points to think about are (1,0,0), (-1,0,0), (0,y,z) along the circle where y^2+z^2=1 and the interior critical point at (0,0,0). What is the value of T in each of those cases?
     
  8. Nov 30, 2011 #7
    T(x,y,z)=60(y^2 + z^2 - x^2)
    T(0,0,0) = 0
    T(1,0,0) = -60
    T(-1,0,0) = -60
    T(0,y,z) = 60(y^2+z^2), but since this is along a circle T(0,y,z) = 60

    Which gives the minimum at (1,0,0) and (-1,0,0), and the max at (0,y,z) where y^2+z^2=1.

    Okay, so from (almost) the beginning:
    λ = -60, x≠0, y=z=0
    λ = 60, x=0

    Since the minima will occur at the smallest λ, and x≠0 provides that λ, but it also requires y=z=0 for that to occur. This leaves T(x,0,0) = -60x2 on -1≤x≤1 because of the unit sphere restriction. x=0 is a critical point, and the end points need to be checked, so T(-1,0,0) = T(1,0,0) = -60, and T(0,0,0)=0 The absolute min occurs at (-1,0,0) and (1,0,0). While (0,0,0) is a relative max (or the absolute max on the interval).

    The maxima occur at x=0 since λ =60 when x=0. This leaves T(0,y,z)=60(y2+z2), with the restriction y2+z2≤1. The maxima are clearly the endpoints, y2+z2=1. So the maximum value for T occurs along that circle, which gives T(0,y,z)=60, where y2+z2=1.

    Hey, it makes sense. Thank you.
     
  9. Nov 30, 2011 #8

    Ray Vickson

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    Of course, this problem is easy to solve without calculus. T = 60(y^2+z^2-x^2) <= 60(y^2+z^2+x^2) <= 60 in the sphere x^2+y^2+z^2<=1. Furthermore, we reach T=60 at any point on y^2+z^2=1, x=0. Also: T >= 60*(-x^2) >= -60, and we reach T=-60 at y=z=0, x=+-1. It is nice to see that the Lagrangian multiplier method dies, indeed, give such results.

    RGV
     
  10. Nov 30, 2011 #9

    HallsofIvy

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    Personally, I wouldn't use "Lagrange Multipliers" for this at all. The temperature function is [itex]T(x,y,z)= 60(y^2+ z^2- x^2)= 60(x^2+ y^2+ z^2- 2x^2)[/itex] which in spherical coordinates is [itex]T(\rho, \theta, \phi)= 60(\rho^2- 2\rho^2sin^2\phi cos^2(\theta))[/itex]. On the surface of the sphere, [itex]\rho= 1[/itex] so that is [itex]T(\theta, \phi)= 60(1- 2sin^2(\phi)cos^2(\theta)[/itex].
     
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