Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

阿Q asked in 科學數學 · 1 decade ago




f(x,y)=x^2+y^2-3xy for (x,y) in R2.

Explain why the function f:R2→R has no local extreme points.

不知道該如何解題才好 所以希望各位大大能幫個忙 非常的急


1 Answer

  • 1 decade ago
    Favorite Answer

    This f function is differentiable everywhere on R2 with

    fx=2x-3y, fy=2y-3x. If this f has a local extrema at a point (a,b) then fx(a,b)=0 and fy(a,b)=0. In fact the system of equations fx=0 and fy=0 has only one solution (a,b)=(0,0).

    To classify this only critical point of f we appy the so-called second derivative test. We calculateD=fxx*fyy-fxy^2=(2)(-3)-0=-6 for every (x,y) hence for (0,0). By the second derivative test, D negative at (a.b) means that (a,b) is just a saddle point, not extremal point of f.

Still have questions? Get your answers by asking now.