find all points where the function has any relative extrema or saddle points and identify the type of…

Explanation:

Step1: Find first - order partial derivatives

The function is (f(x,y)=x^{2}+2y^{2}-xy^{2}). The first - order partial derivatives: (f_{x}=\frac{\partial f}{\partial x}=2x - y^{2}) (f_{y}=\frac{\partial f}{\partial y}=4y-2xy)

Step2: Find critical points

Set (f_{x}=0) and (f_{y}=0). From (f_{y}=4y - 2xy=2y(2 - x)=0), we have two cases: Case 1: (y = 0). Substitute (y = 0) into (f_{x}): (2x-y^{2}=2x=0), so (x = 0). The critical point is ((0,0)). Case 2: (x = 2). Substitute (x = 2) into (f_{x}): (2x-y^{2}=4 - y^{2}=0), then (y=\pm2). The critical points are ((2,2)) and ((2, - 2)).

Step3: Find second - order partial derivatives

(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=2), (f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=-2y), (f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=4 - 2x) The discriminant (D=f_{xx}f_{yy}-(f_{xy})^{2}) For ((0,0)): (f_{xx}(0,0)=2), (f_{xy}(0,0)=0), (f_{yy}(0,0)=4) (D=(2)(4)-0^{2}=8>0) and (f_{xx}(0,0)=2>0). So ((0,0)) is a relative minimum. For ((2,2)): (f_{xx}(2,2)=2), (f_{xy}(2,2)=-4), (f_{yy}(2,2)=4 - 4 = 0) (D=(2)(0)-(-4)^{2}=-16<0). So ((2,2)) is a saddle point. For ((2,-2)): (f_{xx}(2,-2)=2), (f_{xy}(2,-2)=4), (f_{yy}(2,-2)=4 - 4 = 0) (D=(2)(0)-(4)^{2}=-16<0). So ((2,-2)) is a saddle point.

Answer:

A. Relative minimum at ((0,0)) and saddle points at ((2,2)) and ((2, - 2))