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Math Help - xyz^2=5 shortest distance

  1. #1
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    xyz^2=5 shortest distance

    Given the surface xyz^2=5. Prove that on this surface there exists a point closest to the origin, and find that point.

    Justify your solution.

    Thanks alot!
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  2. #2
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    You could try a LaGrange multiplier.

    We want to find the extrema of f(x,y,z)=x^{2}+y^{2}+z^{2} subject to the constraint xyz^{2}=5

    Therefore, we have g(x,y,z)=xyz^{2}

    {\nabla}f(x,y,z)={\lambda}{\nabla}g(x,y,z)

    xi+yj+zk={\lambda}(yz^{2}i+xz^{2}j+2zxyk)

    x={\lambda}yz^{2}; \;\ y={\lambda}xz^{2}; \;\ z={\lambda}2zxy

    {\lambda}=\frac{x}{yz^{2}}...[1]; \;\ {\lambda}=\frac{y}{xz^{2}}...[2]; \;\ {\lambda}=\frac{1}{2xy}...[3]

    Now, you can set [1] and [2] equal and solve for y. Set [1] and [3] equal and solve for z. Then sub them into the constraint to find your x value. Then you can sub that back into the other equations to find y and z.
    Plug those into f(x,y,z) to find the point that is closest the origin.
    There may be several to check.
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  3. #3
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    Quote Originally Posted by galactus View Post
    We want to find the extrema of f(x,y,z)=x^{2}+y^{2}+z^{2} subject to the constraint xyz^{2}=5
    In other words, we want to find the extrema of f(x,y) = x^2+y^2+\frac5{xy}. Put the two partial derivatives equal to zero and you get the equations 2x=5/y^2, 2y=5/x^2. Thus 2x^4=5x. Discounting the solution x=0, you see that x=(5/2)^{1/3} (with y=x and z=2x). I think that's easier than using Lagrange multipliers for this question.
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  4. #4
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    Yes, it is easier, but Lagrange is a good thing to learn.
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  5. #5
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    Well thanks to both of you for your help! It's much appreciated.
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