inequalities (1 Viewer)

=)(= · Jul 10, 2022

Find the minimum value of x^2+y^2 given that xy(x^2-y^2)=x^2+y^2 x not = 0

cossine · Jul 10, 2022

=)(= said:
Find the minimum value of x^2+y^2 given that xy(x^2-y^2)=x^2+y^2 x not = 0

So you have not written coherently.

I am not 100% sure however I think Karush-Kuhn-Tucker condition is what you are after given you have additional constraint x != 0.

Not sure how to solve it otherwise.

5uckerberg · Jul 10, 2022

=)(= said:
Find the minimum value of x^2+y^2 given that xy(x^2-y^2)=x^2+y^2 x not = 0

Is that your question
Find the minimum value of $x^{2}+y^{2}$ given that $xy\left(x^{2}-y^{2}\right)=x^{2}+y^{2}$ , $x\neq{0}$

=)(= · Jul 10, 2022

5uckerberg said:
Is that your question
Find the minimum value of $x^{2}+y^{2}$ given that $xy\left(x^{2}-y^{2}\right)=x^{2}+y^{2}$ , $x\neq{0}$

Yes

cossine · Jul 11, 2022

=)(= said:
Yes

Where did you get the question from?

=)(= · Jul 11, 2022

tutoring

Lith_30 · Jul 11, 2022

I have a slight feeling that this could have something to do with complex numbers, cause if you let $z=x+iy$

$x^2+y^2=|z|^2$ , $\Re(z^2)=x^2-y^2$ and $\frac{1}{2}\Im(z^2)=xy$

so the equation can be $|z|^2=\frac{1}{2}\Im(z^2)\Re(z^2)$ or $|z|^2=\frac{1}{4}\Im(z^4)$

I'm not sure where to go on from there, but maybe it could help you.

EDIT

maybe you can say that $i|z^2|=\frac{1}{4}z^4$ where $\Re(z^4)=0$

That is
$\\(x^2-y^2)^2-4x^2y^2=0\\\\x^4-6x^2y^2+y^4=0\\\\\left(\frac{y}{x}\right)^4-6\left(\frac{y}{x}\right)^2+1=0$
Then
$\left(\frac{y}{x}\right)^2=\frac{6\pm\sqrt{32}}{2}$

cossine · Jul 11, 2022

Lith_30 said:
I have a slight feeling that this could have something to do with complex numbers, cause if you let $z=x+iy$

$x^2+y^2=|z|^2$ , $\Re(z^2)=x^2-y^2$ and $\frac{1}{2}\Im(z^2)=xy$

so the equation can be $|z|^2=\frac{1}{2}\Im(z^2)\Re(z^2)$ or $|z|^2=\frac{1}{4}\Im(z^4)$

I'm not sure where to go on from there, but maybe it could help you.

EDIT

maybe you can say that $i|z^2|=\frac{1}{4}z^4$ where $\Re(z^4)=0$

That is
$\\(x^2-y^2)^2-4x^2y^2=0\\\\x^4-6x^2y^2+y^4=0\\\\\left(\frac{y}{x}\right)^4-6\left(\frac{y}{x}\right)^2+1=0$
Then
$\left(\frac{y}{x}\right)^2=\frac{6\pm\sqrt{32}}{2}$

Sorry how did you get this part:

$i|z^2|=\frac{1}{4}z^4$ where $\Re(z^4)=0$

Lith_30 · Jul 11, 2022

cossine said:
Sorry how did you get this part:

$i|z^2|=\frac{1}{4}z^4$ where $\Re(z^4)=0$

It was an assumption that if $4|z^2|$ was equivalent to the the imaginary component of $z^4$ then you could say $z^4=a+4i|z^2|$ for some real number a.

I just made the number $a$ equal zero cause I wanted the smallest value for $|z^4|$ .

Looking at it now, this is probably wrong. Cause there is no definite reason why $z^4$ has to be purely imaginary.

=)(= · Jul 11, 2022

i used trig sub and i got 1/2 sin(4theta)=1

Trebla · Jul 11, 2022

A bit beyond syllabus but consider a parametric polar form where both r and theta are variables:
$x = r\cos\theta$
$y = r\sin\theta$

This reduces the condition to
$r^2 \sin 4\theta = 4$

Your objective is to find the minimum value of $r^2$ .

Since the RHS is fixed, we minimise $r^2$ by maximising $\sin 4\theta$ .

Hence the minimum value of of $r^2$ is 4.

=)(= · Jul 11, 2022

makes sense thanks

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