Showing a set S is bounded

[cbarker12]

Dear Everybody,

I am having some trouble with proving this set [imath]S=\{(x,y)\in \mathbb{R}^2: 2x^2-4xy+5y^2 \leq 5\}[/imath]. Find any real number [math]R>0[/math] such that [math]\sqrt{x^2+y^2}\leq R[/math] for all [math](x,y)\in S.[/math]
My attempt:
[math]2x^2-4xy+5y^2 =2x^2+(x-y)^2-(x+y)^2+5y^2 \\ \leq x^2+(x-y)-(x+y)+y^2=x^2-2y+y^2 \\ x^2+y^2[/math][math]\sqrt{x^2+y^2}\leq \sqrt{5}[/math].

So [math]R=\sqrt{5}[/math].
Thus S is bounded.

What is the correct technique for elimanating the xy term?
Cbarker12

 

[blamocur]

Do you only need to prove that [imath]S[/imath] is bounded, or are you required to find the exact radius of the bounding circle?

Note that [imath]f(x,y) = 2x^2-4xy+5y^2 = 2(x-y)^2 + y^2[/imath]

 

[pka]

Have you looked at the graph? SEE HERE
Please note that this is the area is the interior of an ellipse not a circle.

 

[Steven G]

I am having some trouble with proving this set S={(x,y)∈R² : 2x²−4xy+5y²≤5}
There is nothing to prove here. S is defined to be {(x,y)∈R² : 2x²−4xy+5y²≤5}

 

[Steven G]

Do you only need to prove that [imath]S[/imath] is bounded, or are you required to find the exact radius of the bounding circle?

Note that [imath]f(x,y) = 2x^2-4xy+5y^2 = 2(x-y)^2 + y^2[/imath]

blamocur meant to say f(x,y) = 2x^2-4xy+5y^2 = 2(x-y)^2 + 3y^2

 

[cbarker12]

Do you only need to prove that [imath]S[/imath] is bounded, or are you required to find the exact radius of the bounding circle?

Note that [imath]f(x,y) = 2x^2-4xy+5y^2 = 2(x-y)^2 + y^2[/imath]

Sorry I need to prove that S is bounded.

 

[cbarker12]

I am having some trouble with proving this set S={(x,y)∈R² : 2x²−4xy+5y²≤5}
There is nothing to prove here. S is defined to be {(x,y)∈R² : 2x²−4xy+5y²≤5}

I forget to add what i need to prove that S is bounded.