Question

Given fxx=6x,fxy=0,fyy=6y, find the nature of stationary point at (-1,2)

Answer 1

Solution:

$f_{xx}=A=6x, \\f_{xy}=B=0, \\f_{yy}=C=6y$

At (-1,2),

$A=6(-1)=-6 \\B=0 \\C=6(2)=12$

Now, we find $AC-B^2$

$=(-6)(12)-0^2 \\=-72<0$

So, nature of stationary point at (-1,2) is a saddle point, neither maximum nor minimum.