Using the Second Partial Derivative Test

Find all local maxima, local minima, and saddle points for \[ z=f(x,y)=x^{2}+xy+y^{2}-6x+6 \]

Solution  The critical points of \(f\) are solutions of the system of equations \[ \left\{ \begin{array}{rcll} f_{x}(x,y)&=&2x+y-6&=0 \\[3pt] f_{y}(x,y)&=&x+2y &=0 \end{array} \right. \]

The only solution is \(x=4\) and \(y=-2,\) so \((4,-2)\) is the only critical point.

The second-order partial derivatives of \(f\) are \[ f_{xx}(x,y)=2\qquad f_{xy}(x,y)=f_{yx}( x,y) =1 \qquad f_{yy}(x,y)=2 \]

The function \(f\) meets the requirements of the Second Partial Derivative test at the critical point \((4,-2)\). Now \[ A=f_{xx}(4,-2)=2\qquad B=f_{xy}(4,-2)=1 \qquad C=f_{yy}(4,-2)=2 \] \[ AC-B^{2}=( 2) ( 2) -1^{2}=3>0 \]

Since \(AC-B^{2}>0\) and \(A= f_{xx}(4,-2)=2>0,\) the Second Partial Derivative Test guarantees that \(f\) has a local minimum at \((4,-2)\). The local minimum value is \(z=f(4,-2)=-6\). See Figure 16.