Fastest Method for 8767 q55

Forum for the GRE subject test in mathematics.
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yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Fastest Method for 8767 q55

Post by yoyobarn » Sun Jan 08, 2012 10:18 am

Q55. Let f(x,y)=x^3+y^3+3xy for all real x and y. Then there exist distinct points P and Q such that f has a...

Is there a faster method other than using the "2nd derivative test"?

Usual method:
Solve $$f_x=3x^2+3y=f_y=3y^2+3x$$ to get (x,y)=0 or (x,y)=(-1,-1).

Then check the sign of D=36xy-9, and $$f_{xx}=6x$$

For x=0, D<0, so it is saddle point.

For x=-1, D>0 and $$f_{xx}=6x<0$$ so it is a maximum.


Just curious if there is a faster and easier way out for this question?

Thank you very much.

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: Fastest Method for 8767 q55

Post by owlpride » Sun Jan 08, 2012 10:18 pm

Computing the Hessian is fool-proof and takes less than a minute. I doubt there's a faster reliable method out there.

For some nice functions you could figure it out "by hand" but looking for the right approach might take you longer than computing 4 partials. For example, it's immediately obvious that (0,0) is a critical point, and not a local min or max because f(x,0) is negative for negative x and positive for positive x.

If I had 10 seconds left on the exam and knew that (-1,-1) is a critical point, I would guess that (-1,-1) is a local max because f(-1,-1) > f(0,0) and f(-x,-x) goes to negative infinity as x gets large; and it's also easy to check that f(x,y) < f(-1,-1) for a few other points on the integer lattice near (-1,-1). That doesn't prove it, but it's a strong indication for a function as simple as a cubic.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: Fastest Method for 8767 q55

Post by yoyobarn » Sun Jan 08, 2012 11:51 pm

Thanks a lot.



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