Question

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

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) \((1,-3)\)