# maximum/minimum

#### Dinkydoe

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?

#### ojones

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 $$\displaystyle x=y^2+z^2$$?

#### Dinkydoe

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
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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• Dinkydoe

#### HallsofIvy

MHF Helper
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.

• Dinkydoe

#### Dinkydoe

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