Dec 2009
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?
May 2010
Los Angeles, California
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 \(\displaystyle x=y^2+z^2\)?
Dec 2009
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.


MHF Hall of Honor
Jan 2009
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 \(\displaystyle x + \frac{1}{4} = \frac{1}{\sqrt{2}} \frac{3}{2\sqrt{2}} \cos{t} \)

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

\(\displaystyle x \leq \frac{3}{4} - \frac{1}{4} = \frac{1}{2} \)

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 \(\displaystyle (\frac{1}{2})^2 + (\frac{1}{2}) = \frac{3}{4}\) .
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MHF Helper
Apr 2005
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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Dec 2009
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 \(\displaystyle x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\frac{1}{2}]\)