# Math Help - maximum/minimum

1. ## 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?

2. This looks like a Lagrange multiplier problem. Have you heard of this method?

For $C$ did you mean to square the $x$ in $x=y^2+z^2$?

3. 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.

4. 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
$

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

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 $(\frac{1}{2})^2 + (\frac{1}{2}) = \frac{3}{4}$ .

5. 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.

6. 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 $x\in [-\frac{1}{2}+\frac{1}{2}\sqrt{3},\frac{1}{2}]$