# Thread: Minimum point of f(x,y)

1. ## Minimum point of f(x,y)

I have the following function: $f(x,y)=x^4+2x^3y+x^2y^2+y^4$ and I know that there is a critical point at $(0,0)$. I have been asked to determined the type of critical point. The second derivatives give no information, and I'm not supposed to consider higher derivatives.

"Obviously" it is a minimum point, so I thought it would be enough to show that $f(x,y)\geq 0$ near $(0,0)$.

By introducing polar coordinates I got:

$f(r\cos(\theta),r\sin(\theta))=...=r^4\left[1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\right]$.

Since $r^4\geq 0$, it should be enough to show that $1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\geq 0$, but I seem unable to do it. Does this seem like the right method, and if so, does anyone know how to go on from here?

Another thought I had was to compare with $(x+y)^4$ or $(x-y)^4$, but that didn't get me anywhere.

Aliquantus

2. Originally Posted by Aliquantus
I have the following function: $f(x,y)=x^4+2x^3y+x^2y^2+y^4$ and I know that there is a critical point at $(0,0)$. I have been asked to determined the type of critical point. The second derivatives give no information, and I'm not supposed to consider higher derivatives.

"Obviously" it is a minimum point, so I thought it would be enough to show that $f(x,y)\geq 0$ near $(0,0)$.

By introducing polar coordinates I got:

$f(r\cos(\theta),r\sin(\theta))=...=r^4\left[1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\right]$.

Since $r^4\geq 0$, it should be enough to show that $1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\geq 0$, but I seem unable to do it. Does this seem like the right method, and if so, does anyone know how to go on from here?

Another thought I had was to compare with $(x+y)^4$ or $(x-y)^4$, but that didn't get me anywhere.

Aliquantus
Note the following

$x^4 + 2x^3y+x^2y^2+y^4 = x^2\left(x^2+2xy+y^2\right) + y^4 = x^2\left(x+y\right)^2+y^4$.

3. Of course, thanks alot!