# Another local extreme problem

• Oct 13th 2012, 12:02 PM
MajMinor
Another local extreme problem
I am facing another problem that I can't compute the extremes for:(Doh)

f(xy)=ex(x2-y2)

so far I have:

fx= 2xex + ex(x2-y2)
fy= -2yex
fxx=2ex+2xex + ex(x2-y2) + 2xex
fyy= -2ex
fxy= -2yex

How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.

Thanks for any help.
• Oct 13th 2012, 12:13 PM
TheEmptySet
Re: Another local extreme problem
Quote:

Originally Posted by MajMinor
I am facing another problem that I can't compute the extremes for:(Doh)

f(xy)=ex(x2-y2)

so far I have:

fx= 2xex + ex(x2-y2)
fy= -2yex
fxx=2ex+2xex + ex(x2-y2) + 2xex
fyy= -2ex
fxy= -2yex

How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.

Thanks for any help.

If you set the partial with respect to x equal to 0 you get

$\displaystyle 2xe^{x}+e^{x}(x^2-y^2)=0 \\ e^{x}(2x+x^2-y^2)=0$

Since the exponential is never zero his gives

$\displaystyle 2x+x^2-y^2=0 \iff y=\pm \sqrt{x^2+2x}$

If we plug this into the 2nd equation we get

$\displaystyle 2(\pm \sqrt{x^2+2x})e^{x}=0$

Again the exponential can never be zero so we get

$\displaystyle (\pm \sqrt{x^2+2x})=0 \iff x^2+2x=0^2 \iff x(x+2)=0$

So we get the two solutions

$\displaystyle x=0 \quad x=-2$

Can you finish from here?
• Oct 13th 2012, 12:26 PM
MajMinor
Re: Another local extreme problem
Yes. Thank you very much! It's the little things that always trip me up.