unconstrained optimization: f(x,y) = x/(1+x^2+y^2)

[woolley]

Find the local extreme values and classify each as maxima, minima, or neither.

f(x,y) = x/(1+x^2+y^2)

Here is what I have done:

Df = gradient of f = (1-x^2+y^2)/[(1+x^2+y^2)^2]

(-2xy)/[(1+x^2+y^2)^2]

I set them equal to zero, then solved the equations, and got extreme points (1,0) and (0,1).

Then I set up Hessian (D^2f) and the first one (with respect to x twice) is ridiculously long. So, I thought maybe I went wrong somewhere.

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[soroban]

Hello, woolley!

Find the local extreme values and classify each as maxima, minima, or neither.

. . \(\displaystyle f(x,y) \:= \:\frac{x}{1\,+\,x^2\,+\,y^2}\)

Here is what I have done:

. . \(\displaystyle \frac{\partial f}{\partial x}\:=\:\frac{1\,-\,x^2\,+\,y^2}{(1\,+\,x^2\,+\,y^2)^2}\)

. . \(\displaystyle \frac{\partial f}{\partial y} \:=\:\frac{-2xy}{(1\,+\,x^2\,+\,y^2)^2}\)


I set them equal to zero, then solved the equations,
. . and got extreme points (1,0) and (0,1). . . . . no

We have: \(\displaystyle \:1\,-\,x^2\,+\,y^2\:=\:0\;\;\Rightarrow\;\;x^2\,-\,y^2\:=\:1\;\) [1]
. . . .and: \(\displaystyle \:-2xy \:=\:0\;\;\Rightarrow\;\;xy \:=\:0\;\) [2]


From [2]: \(\displaystyle x\,=\,0 \text{ or }y\,=\,0\)

. . If \(\displaystyle x\,=\,0\), then [1] becomes: \(\displaystyle \,y^2\,=\,-1\) . . . no real roots

. . If \(\displaystyle y\,=\,0\), then [1] becomes: \(\displaystyle \,x^2\,=\,1\;\;\Rightarrow\;\;x\,=\,\pm1\)

The extreme points are: \(\displaystyle \:\fbox{\,(1,\,0),\;(-1,\,0)\,}\)


The Second Partials Test is very messy . . . I'm still working on it.

 

[woolley]

what a mess indeed!

OK, this is what I got...wish I could use TeX!

w.r. to x twice:

(2x^3-6xy^2-6x)/(1+x^2+y^2)^3

w.r. to x then y:

(2x^2y-2y-2y^3)/(1+x^2+y^2)^3

w.r. to y then x:

(4x^2y-2y-4y^3)/(1+x^2+y^2)^3

w.r. to y twice:

(8xy^2-4x^2-4x^4-4x^2y^2)/(1+x^2+y^2)^3