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Math Help - maximum/minimum

  1. #1
    Senior Member Dinkydoe's Avatar
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    maximum/minimum

    I feel my IQ drops with every second I look at this excercise, that's why I ask you guys:

    Given C\subset \mathbb{R}^3 with (x,y,z) that satisfy 2x^2+2y^2+z^2=1 and x=y^2+z^2

    Find (x,y,z)\in C such that the distance to the origin is maximal/minimal

    However, I can't find points that satisfy to both equations of C

    We can derive
    (1) x\geq 0
    (2) y^2=1-2x^2-x\geq 0
    (3) z^2=2x^2+2x-1\geq 0

    From (2) we get x\in [0,\frac{1}{2}]
    From (3) we get x\geq \frac{1}{2}

    Only x= \frac{1}{2} seems ok. But it gives y^2=z^2=0, ...scheisse

    So, I can't find any x that could possibly satisfy the equations, let alone y,z

    Can someone fix my brains? What's wrong here?
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  2. #2
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    This looks like a Lagrange multiplier problem. Have you heard of this method?

    For C did you mean to square the x in x=y^2+z^2?
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Yes, I'm familiar with the method. And the excercise itself is not my problem.

    My problem is the definition of C, with x=y^2+z^2, from wich I derive that C is empty

    So I think there's a mistake in the excercise. I think indeed it must be x^2=y^2+z^2.

    But no, I didn't mean to square the x, this is exactly the excercise.
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  4. #4
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    We can first eliminate  z
    From  x = y^2 + z^2

     z^2 = x - y^2


     2x^2 + 2y^2 + z^2 = 1
     2x^2 + 2y^2  + x - y^2 = 1

     2x^2 + x + y^2 = 1

    We have

     D^2 = x^2 + y^2 + z^2 = x^2 + x = ( x + \frac{1}{2} )^2  - \frac{1}{4}

    Then we need to find the range of  x

     2x^2 + x + y^2 = 1 <br />

     2(x + \frac{1}{4} )^2 + y^2 = 1 + \frac{1}{8} = \frac{9}{8}


    Let  x + \frac{1}{4} = \frac{1}{\sqrt{2}} \frac{3}{2\sqrt{2}} \cos{t}

     y = \frac{3}{2\sqrt{2}} \sin{t}

    so
     x \leq \frac{3}{4} - \frac{1}{4} = \frac{1}{2}

    Since the quadratic function  x^2 + x  is increasing for  x \geq 0  > - \frac{1}{2} , we have the max. of  D^2 is  (\frac{1}{2})^2 + (\frac{1}{2}) =  \frac{3}{4} .
    Last edited by simplependulum; May 17th 2010 at 03:05 AM.
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  5. #5
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    Why should C be empty? C consists of points satisfying both 2x^2+ 2y^2+ z^2= 1, an ellipse, and x= y^2+ z^2, a parboloid with axis along the x- axis. Certainly those intersect.

    We can write the first equation as 2x^2+ 2y^2+ 2z^2- z^2= 2x^2+ 2(y^2+ z^2)+ z^2 = 2x^2+ 2x+ z^2= 2(x^2+ x+ 1/4- 1/4)+ 2z^2= 2(x+ 1/2)^2+ z^2= 1, so the two surfaces intersect on an ellipse.
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  6. #6
    Senior Member Dinkydoe's Avatar
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    Yeah, thank you both.

    Needed my brain to get fixed, perhaps I wasn't thinking clear.

    But the mistake I made was at (3). Somehow I thought the positive root was x=\frac{1}{2} wich is clearly wrong.

    getting the positive root of 2x^2+2x-1 is x_0 =-\frac{1}{2}+\frac{1}{2}\sqrt{3}. This gives x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\infty)

    Hence we need to take the intersection of (1),(2) and we find x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\frac{1}{2}]

    I guess, no-one really did an effort to see that I made a lame calculation error at (3). Now clearly C is not empty and consists of the points

    (x, 1-2x^2-x, 2x^2+2x-1) with x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\frac{1}{2}]
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