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Thread: Nearest and farthest point

  1. #1
    Oct 2008

    Nearest and farthest point

    The plane $\displaystyle x+y+2z=2$ intersects the paraboloid $\displaystyle z=x^2+y^2$ in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

    What I did:

    Let $\displaystyle f(x,y)=x+y+2z-2$ and $\displaystyle g(x,y)=z-x^2-y^2$
    $\displaystyle f_x=1,f_y=1,f_z=2$ and $\displaystyle g_x=-2x, g_y=-2y,g_z=1$


    $\displaystyle 1=-2x\lambda$
    $\displaystyle 1=-2y\lambda$
    $\displaystyle 2=\lambda$
    $\displaystyle z=x^2+y^2$

    $\displaystyle x=-1/4,y=-1/4,z=1/8$

    But this is none of the answer. The answers are $\displaystyle (1/2,1/2,1/2) and (-1,-1,2)$

    Can someone point out my mistake?
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009
    Here you have 2 constraints

    x + y + 2z = 2 call it f

    z-x^2 -y^2 = 0 call it g

    You want to maxmimize the distance to the origin
    (maximize the square of the distance)

    Call it D = x^2 + y^2 + z^2

    gradD= L1 gradf + L2 gradg

    This leads to the equations:

    1) 2x = L1 -2xL2
    2) 2y = L1 -2yL2
    3) 2z= 2L1 + L2
    4) x + y + 2z =2
    5) z = x^2 + y^2

    i Eliminate L2 between eqns 1) and 2) and you obtain x = y

    ii Use this in 4) to get z = 1 - x

    iii Use this in 5 to get x = 1/2 or -1

    Use this to finish
    Last edited by Calculus26; Nov 18th 2009 at 11:12 AM.
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