Given fxx=6x,fxy=0,fyy=6y, find the nature of stationary point at (-1,2)
fxx=A=6x,fxy=B=0,fyy=C=6yf_{xx}=A=6x, \\f_{xy}=B=0, \\f_{yy}=C=6yfxx=A=6x,fxy=B=0,fyy=C=6y
At (-1,2),
A=6(−1)=−6B=0C=6(2)=12A=6(-1)=-6 \\B=0 \\C=6(2)=12A=6(−1)=−6B=0C=6(2)=12
Now, we find AC−B2AC-B^2AC−B2
=(−6)(12)−02=−72<0=(-6)(12)-0^2 \\=-72<0=(−6)(12)−02=−72<0
So, nature of stationary point at (-1,2) is a saddle point, neither maximum nor minimum.
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