Find a point where f(x,y) = x^2 + 4xy + 3y^2 < 0

[Cade]

Homework Statement

Find a point where f(x,y) = x^2 + 4xy + 3y^2 < 0

Homework Equations

Turning points are where df/dx = 0 and df/dy = 0

The Attempt at a Solution

df = (2x + 4y)dx + (4x + 6y)dy

2x + 2y = 0, 4x + 6y = 0 -> turning point at (0,0)

From this step, how can I find a point (x,y) where f(x,y) is negative?

 

[Dick]

Why don't you try and complete the square? That should make it easy to find a point by just looking at it.

 

[Cade]

Completing the square gives me
f(x,y) = (x + 2y)^2 - y2

How should I continue?

 

[Dick]

Completing the square gives me
f(x,y) = (x - 2y)^2 - y2

How should I continue?

Pick a point where x-2y=0 and y is nonzero, right? Except I get (x+2y)^2-y^2.

 

[Cade]

Edit: Oh whoops, my equation was wrong.

Oh, I see. Thanks for your help. :)

x = 2, y = -1

 

[Dick]

Oh, I see. Thanks for your help. :)

x = 2, y = 1

I think you got the square completed a bit wrong though, try it again.

 

[Cade]

Yes, I see that now. I mistakenly wrote the wrong sign.

 

[epenguin]

Rather a strange question if I understand it. Don't points satisfying that inequality fill up an area (I think it is 1/12) of the infinite x,y plane? And you are asked to find only one?

I would factorise the expression, draw lines x^2 + 4xy + 3y^2 = 0 and then see.

 

[Cade]

I was only asked to find a single point. This question came up out of context, so I didn't know how to attempt it.

 

[HallsofIvy]

$(x+ 2y)^2- y^2= 0$ when $y= x+ 2y$ or $y= -(x+ 2y)$ which reduce to $y= -x$ and $3y= -x$, two lines that intersect at (0,0). They divide the plane into four regions. Since the given function is a polynomial in x and y, it can only change from ">" to "<", and vice-versa, where we have "=". That is, if a single point in one of those four regions gives ">" then all do. Check one point in each region to see which is true for that region.