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Math Help - calc 3, local maxes and mins

  1. #1
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    Question calc 3, local maxes and mins

    find the points on the ellipsoid x^2 + (y^2)/4 + (z^2)/9 =1
    that are closest to the origin


    i'm not sure how to go about solving this problem. it has to do with maxes or mins i'm guessing since that's what we just learned. anyone know what i should do?

    i tried taking the derivative and setting it equal to zero and solving for x, y, and z but that gives me (0,0,0) and the answer in the back is (+ or - 1, 0,0)
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by holly123 View Post
    find the points on the ellipsoid x^2 + (y^2)/4 + (z^2)/9 =1
    that are closest to the origin


    i'm not sure how to go about solving this problem. it has to do with maxes or mins i'm guessing since that's what we just learned. anyone know what i should do?

    i tried taking the derivative and setting it equal to zero and solving for x, y, and z but that gives me (0,0,0) and the answer in the back is (+ or - 1, 0,0)
    You are trying to minimize the function f(x,y,z)=\sqrt{x^2+y^2+z^2} (the regular Euclidean norm on \mathbb{R}^3) subject to the constraint x^2+\frac{y^2}{4}+\frac{z^2}{9}=1. However, minimizing f is the same as minimizing x^2+y^2+z^2. (Do you see why?)

    I would use Lagrange multipliers:

    \nabla_f=\langle 2x,2y,2z\rangle=\lambda\nabla_{con}=\left\langle 2\lambda x,\frac{\lambda}{2}y,\frac{2\lambda}{9}z\right\ran  gle

    So you need to solve the system:

    \left\{\begin{array}{l}2x=2\lambda x\\2y=\frac{\lambda}{2}y\\2z=\frac{2\lambda}{9}z\\  x^2+\frac{y^2}{4}+\frac{z^2}{9}=1\end{array}\right  \}

    This system is not nearly as hard as it looks. Note that if \lambda=0, then x,y,z=0 as well (from the first three equations), but this contradicts the fourth equation, because we'd get 0=1. So \lambda\neq0. But from the second and third equations, this implies that y,z=0. Now our system is simpler:

    \left\{\begin{array}{l}2x=2\lambda x\\x^2=1\end{array}\right\}

    Solving this is easy. We get \lambda=1 (which is irrelevant for this problem) and x=\pm1.

    So the points closest to the origin are (1,0,0) and (-1,0,0), just like the answer key says.

    If you think about this geometrically, it makes sense. The points closest to the origin will be on the minor axis of the ellipsoid, which in this case is the x-axis, and can be obtained by plugging in 0 for y,z and solving for x.
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