Find all local maxima, local minima, and saddle points for $$z=f(x,y)=x^{3}+y^{2}+2xy-4x-3y+5$$

Solution  The critical points are the solutions of the system of equations $$\left\{ \begin{array}{rcll} f_{x}(x,y)&=&3x^{2}+2y-4&=0 \\[3pt] f_{y}(x,y)&=&2y+2x-3&=0 \end{array}\right.$$

We solve the system using substitution. Solving for $y$ in $f_{y}(x,y)=0$, we obtain $y=\dfrac{1}{2}(3-2x)$. Then if we substitute for $y$ in $f_{x}(x,y)=0$, the result is $$\begin{eqnarray*} 3x^{2}+2\,\left[ \dfrac{1}{2}(3-2x)\right] -4& =0 \\[4pt] 3x^{2}-2x-1& =0 \\[4pt] (3x+1)(x-1)& =0 \\[4pt] x=-\dfrac{1}{3}\qquad \hbox{or }\qquad x& =1 \end{eqnarray*}$$

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When $x=-\dfrac{1}{3}$, then $y=\dfrac{1}{2}(3-2x)=\dfrac{1}{2}\left( 3+ \dfrac{2}{3}\right) =\dfrac{11}{6}$.

When $x=1$, then $y=\dfrac{1}{2}(3-2x)=\dfrac{1}{2}$.

The critical points are $\left( -\dfrac{1}{3},\dfrac{11}{6}\right)$ and $\left( 1,\dfrac{1}{2}\right)$.

The second-order partial derivatives of $f$ are $$f_{xx}(x,y)=6x\qquad f_{xy}(x,y)=2\qquad f_{yy}(x,y)=2$$

Figure 17 $$\begin{array}{l} z=f(x,y)\\ \,\,\,\, =x^{3}+y^{2}+2xy-4x-3y+5 \\ \end{array}$$

The function $f$ satisfies the requirements of the Second Partial Derivative test at each critical point. At $\left( -\dfrac{1}{3},\dfrac{11}{6}\right)$, $$\begin{eqnarray*} &A=f_{xx}\left( -\dfrac{1}{3},\dfrac{11}{6}\right) =-2\qquad B=f_{xy}\left( -\dfrac{1}{3},\dfrac{11}{6}\right) =2\qquad C=f_{yy}\left( -\dfrac{1}{3},\dfrac{11}{6}\right) =2&\\[4pt] &AC-B^{2}=-4-4=-8<0& \end{eqnarray*}$$

So, $\left( -\dfrac{1}{3},\dfrac{11}{6},\dfrac{317}{108}\right)$ is a saddle point of $f$. At $\left( 1,\dfrac{1}{2}\right)$, $$A=f_{xx}\left( 1,\dfrac{1}{2}\right) =6\qquad B=f_{xy}\left( 1,\dfrac{1}{2}\right) =2\qquad C=f_{yy}\left( 1,\dfrac{1}{2}\right)=2$$

Since $AC-B^{2}=8>0$, $f$ has a local minimum at $\left( 1,\dfrac{1}{2}\right)$, where $f\left( 1,\dfrac{1}{2}\right) =\dfrac{7}{4}$