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

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

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

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

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

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

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

    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 $\displaystyle C$ did you mean to square the $\displaystyle x$ in $\displaystyle x=y^2+z^2$?
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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 $\displaystyle C$, with $\displaystyle x=y^2+z^2$, from wich I derive that $\displaystyle C$ is empty

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

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

    $\displaystyle z^2 = x - y^2 $


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

    $\displaystyle 2x^2 + x + y^2 = 1 $

    We have

    $\displaystyle 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 $\displaystyle x $

    $\displaystyle 2x^2 + x + y^2 = 1
    $

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


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

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

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

    Since the quadratic function $\displaystyle x^2 + x $ is increasing for $\displaystyle x \geq 0 > - \frac{1}{2} $ , we have the max. of $\displaystyle D^2 $ is $\displaystyle (\frac{1}{2})^2 + (\frac{1}{2}) = \frac{3}{4}$ .
    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 $\displaystyle 2x^2+ 2y^2+ z^2= 1$, an ellipse, and $\displaystyle x= y^2+ z^2$, a parboloid with axis along the x- axis. Certainly those intersect.

    We can write the first equation as $\displaystyle 2x^2+ 2y^2+ 2z^2- z^2= 2x^2+ 2(y^2+ z^2)+ z^2$$\displaystyle = 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 $\displaystyle x=\frac{1}{2}$ wich is clearly wrong.

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

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

    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

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