MHB 243.14.7.5 Find all local extreme values

[karush]

$\tiny{243.14.7.5}$

$\textsf{Find all local extreme values of the given function and identify each}$
$\textsf{as a local maximum,local minimum,or saddlepoint}$
$\textit{$f$ has a} \textbf{ saddle point } \textit{at $(a,b)$ if}$
\begin{align*} \displaystyle
f_{xx}f_{yy}-fxy^2&\ge 0 \text{ at } (a,b)
\end{align*}
\begin{align*} \displaystyle
f_5(x,y)&=2xy-3x+3y\\
f_x&=2y-3\\
f_y&=2x+3\\
f_x(3/2)&=2y-3=0\\
f_y(-3/2)&=2x+3=0\\
f\left[-\frac{3}{2},\frac{3}{2} \right]
&=2\left[-\frac{3}{2} \cdot \frac{3}{2} \right]
-3\left[-\frac{3}{2} \right]
+3\left[\frac{3}{2} \right]=\frac{9}{2}\\
f_{xx}&=2\\
f_{yy}&=2\\
f_{xx}f_{yy}&=2 \cdot 2-[2 \cdot 2]^2\\
&=4-16=-12 < 0
\end{align*}
\begin{align*} \displaystyle
\textit{Book Answer: }
f\left[-\frac{3}{2},\frac{3}{2} \right]&=\color{red}{\frac{9}{2}}
\textit{ ,Saddle point}
\end{align*}first time
so suggestions?

 

[MarkFL]

We are given:

$$f(x,y)=2xy-3x+3y$$

Identify the critical point(s):

$$f_x(x,y)=2y-3=0\implies y=\frac{3}{2}$$

$$f_y(x,y)=2x+3=0\implies x=-\frac{3}{2}$$

And so our critical point is:

$$(a,b)=\left(-\frac{3}{2},\frac{3}{2}\right)$$

Now, according to the second partials test, we let:

$$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left[f_{xy}(x,y)\right]^2$$

And we find:

$$D(a,b)=0\cdot0-(2)^2=-4<0$$

Hence, we conclude that $f(a,b)$ is not an extremum (it is a saddle point).