maximum/minimum

Dec 2009
411
131
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
274
67
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
411
131
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.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
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 \(\displaystyle (\frac{1}{2})^2 + (\frac{1}{2}) = \frac{3}{4}\) .
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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
411
131
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 \(\displaystyle x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\frac{1}{2}]\)