- #1

cbarker1

Gold Member

MHB

- 349

- 23

- TL;DR Summary
- Showing a set is bounded.

Dear Everybody,

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:

##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##

##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.

Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks

cbarker1

I am having some trouble with proving this set ##S=\{(x,y)\in \mathbb{R}^2: 3x^2-4xy+5y^2 \leq 5\}## is bounded. Find a real number ##R>0## such that ##\sqrt{x^2+y^2}\leq ## for all ##(x,y)\in S.##

My attempt:

##3x^2-4xy+5y^2 =3x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \leq x^2+y^2##

##\sqrt{x^2+y^2}\leq \sqrt{5}##.

So ##R=\sqrt{5}##.

Thus S is bounded.

What is the correct technique for elimanating the xy term?

Thanks

cbarker1

Last edited: