consider the following.\n$f(x,y)=x^{3}+y^{3}-3x^{2}+9y^{2}+3x+27y+26$\n(a) find the critical…

consider the following.\n$f(x,y)=x^{3}+y^{3}-3x^{2}+9y^{2}+3x+27y+26$\n(a) find the critical point.\n$(x,y)=$( )\n(b) test for relative extrema.\n$\\bigcirc$ the critical point is an absolute maximum.\n$\\bigcirc$ the critical point is an absolute minimum.\n$\\bigcirc$ the critical point is a saddle point.\n(c) list the critical points for which the second partials test fails.\n$(x,y)=$( )

consider the following.\n$f(x,y)=x^{3}+y^{3}-3x^{2}+9y^{2}+3x+27y+26$\n(a) find the critical point.\n$(x,y)=$( )\n(b) test for relative extrema.\n$\\bigcirc$ the critical point is an absolute maximum.\n$\\bigcirc$ the critical point is an absolute minimum.\n$\\bigcirc$ the critical point is a saddle point.\n(c) list the critical points for which the second partials test fails.\n$(x,y)=$( )

Answer

Explanation:

Step1: Find the first - order partial derivatives

The first - order partial derivatives of (f(x,y)=x^{3}+y^{3}-3x^{2}+9y^{2}+3x + 27y+26) are: (f_{x}=\frac{\partial f}{\partial x}=3x^{2}-6x + 3) and (f_{y}=\frac{\partial f}{\partial y}=3y^{2}+18y + 27). Set (f_{x}=0) and (f_{y}=0): For (f_{x}=3x^{2}-6x + 3 = 3(x^{2}-2x + 1)=3(x - 1)^{2}=0), we get (x = 1). For (f_{y}=3y^{2}+18y + 27=3(y^{2}+6y + 9)=3(y + 3)^{2}=0), we get (y=-3). So the critical point ((x,y)=(1,-3)).

Step2: Find the second - order partial derivatives

(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=6x-6), (f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=0), (f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=6y + 18). The discriminant (D=f_{xx}f_{yy}-(f_{xy})^{2}=(6x - 6)(6y + 18)). At the critical point ((1,-3)), (f_{xx}(1,-3)=6\times1-6 = 0), (f_{yy}(1,-3)=6\times(-3)+18 = 0), (D=(6\times1 - 6)(6\times(-3)+18)=0).

Answer:

(a) ((1,-3)) (c) ((1,-3))