Find the critical point(s)
∂x∂f=4y2+3y
∂y∂f=8xy+3x Then
⎩⎨⎧∂x∂f=0∂y∂f=0
{4y2+3y=08xy+3x=0Critical points (0,0),(0,−3/4)
Point (0,0)
∂x2∂2f(0,0)=0
∂y2∂2f(0,0)=8(0)=0
∂x∂y∂2f(0,0)=∂y∂x∂2f(0,0)=8(0)+3=3
D(0,0)=∣∣0330∣∣=−9<0Then the f(0,0) is not a local maximum or minimum.
The critical point (0,0) is a saddle point.
Point (0,−3/4)
∂x2∂2f(0,−3/4)=0
∂y2∂2f(0,−3/4)=8(0)=0
∂x∂y∂2f(0,−3/4)=∂y∂x∂2f(0,−3/4)
=8(−3/4)+3=−3
D(0,0)=∣∣0−3−30∣∣=−9<0Then the f(0,−3/4) is not a local maximum or minimum.
The critical point (0,−3/4) is a saddle point.
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