3 The nature of a stationary point
We state, without proof, a relatively simple test to determine the nature of a stationary point, once located. If the surface is very flat near the stationary point then the test will not be sensitive enough to determine the nature of the point. The test is dependent upon the values of the second order derivatives: ${f}_{xx},{f}_{yy},{f}_{xy}$ and also upon a combination of second order derivatives denoted by $D$ where
$\phantom{\rule{2em}{0ex}}D\equiv \frac{{\partial}^{2}f}{\partial {x}^{2}}\times \frac{{\partial}^{2}f}{\partial {y}^{2}}{\left(\frac{{\partial}^{2}f}{\partial x\partial y}\right)}^{2}$ , which is also expressible as $D\equiv {f}_{xx}{f}_{yy}{\left({f}_{xy}\right)}^{2}$
The test is as follows:
Key Point 4
Test to Determine the Nature of Stationary Points
 At each stationary point work out the three second order partial derivatives.

Calculate the value of
$D={f}_{xx}{f}_{yy}{\left({f}_{xy}\right)}^{2}$
at each stationary point.
Then, test each stationary point in turn:

If
$D<0$
the stationary point is a
saddle point
.
If $D>0$ and $\frac{{\partial}^{2}f}{\partial {x}^{2}}>0$ the stationary point is a local minimum .
If $D>0$ and $\frac{{\partial}^{2}f}{\partial {x}^{2}}<0$ the stationary point is a local maximum.
If $D=0$ then the test is inconclusive (we need an alternative test).
Example 11
The function: $f\left(x,y\right)={x}^{4}+{y}^{4}36xy$ has stationary points at $\left(0,0\right),\phantom{\rule{1em}{0ex}}\left(3,3\right),\phantom{\rule{1em}{0ex}}\left(3,3\right).$ Use Key Point 4 to determine the nature of each stationary point.
Solution
We have $\frac{\partial f}{\partial x}={f}_{x}=4{x}^{3}36y$ and $\frac{\partial f}{\partial y}={f}_{y}=4{y}^{3}36x$ .
Then $\frac{{\partial}^{2}f}{\partial {x}^{2}}={f}_{xx}=12{x}^{2},\phantom{\rule{1em}{0ex}}\frac{{\partial}^{2}f}{\partial {y}^{2}}={f}_{yy}=12{y}^{2}$ , $\frac{{\partial}^{2}f}{\partial x\partial y}={f}_{yx}=36$ .
A tabular presentation is useful for calculating $D={f}_{xx}{f}_{yy}{\left({f}_{xy}\right)}^{2}$ :
Point  Point  Point  
Derivatives  $\left(0,0\right)$  $\left(3,3\right)$  $\left(3,3\right)$ 
${f}_{xx}$  $0$  $108$  $108$ 
${f}_{yy}$  $0$  $108$  $108$ 
${f}_{xy}$  $36$  $36$  $36$ 
$D$  $<0$  $>0$  $>0$ 
$\left(0,0\right)$ is a saddle point; $\left(3,3\right)$ and $\left(3,3\right)$ are both local minima.
Task!
Determine the nature of the stationary points of $f\left(x,y\right)={x}^{3}+{y}^{2}3x6y1$ , which are $\left(1,3\right)$ and $\left(1,3\right)$ .
Write down the three second partial derivatives:
${f}_{xx}=6x,\phantom{\rule{1em}{0ex}}{f}_{yy}=2,\phantom{\rule{1em}{0ex}}{f}_{xy}=0.$
Now complete the table below and determine the nature of the stationary points:
Point  Point  
Derivatives  $\left(1,3\right)$  $\left(1,3\right)$ 
${f}_{xx}$  $6$  $6$ 
${f}_{yy}$  $2$  $2$ 
${f}_{xy}$  $0$  $0$ 
$D$  $>0$  $<0$ 
State the nature of each stationary point:
$\left(1,3\right)$ is a local minimum; $\left(1,3\right)$ is a saddle point.
For most functions the procedures described above enable us to distinguish between the various types of stationary point. However, note the following example, in which these procedures fail.
Given $f\left(x,y\right)={x}^{4}+{y}^{4}+2{x}^{2}{y}^{2}.$
$\phantom{\rule{2em}{0ex}}\frac{\partial f}{\partial x}=4{x}^{3}+4x{y}^{2},\phantom{\rule{2em}{0ex}}\frac{\partial f}{\partial y}=4{y}^{3}+4{x}^{2}y,$
$\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}f}{\partial {x}^{2}}=12{x}^{2}+4{y}^{2},\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}f}{\partial {y}^{2}}=12{y}^{2}+4{x}^{2},\phantom{\rule{2em}{0ex}}\frac{{\partial}^{2}f}{\partial x\partial y}=8xy$
Location
: The stationary points are located where
$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0,$
that is,
where
$4{x}^{3}+4x{y}^{2}=0$
and
$4{y}^{3}+4{x}^{2}y=0$
. A simple
factorisation implies
$4x\left({x}^{2}+{y}^{2}\right)=0$
and
$4y\left({y}^{2}+{x}^{2}\right)=0$
.
The only solution which satisfies both equations is
$x=y=0$
and therefore the only
stationary point is
$\left(0,0\right)$
.
Nature : Unfortunately, all the second partial derivatives are zero at $\left(0,0\right)$ and therefore $D=0$ , so the test, as described in Key Point 4, fails to give us the necessary information.
However, in this example it is easy to see that the stationary point is in fact a local minimum. This could be confirmed by using a computer generated graph of the surface near the point $\left(0,0\right)$ . Alternatively, we observe ${x}^{4}+{y}^{4}+2{x}^{2}{y}^{2}\equiv {\left({x}^{2}+{y}^{2}\right)}^{2}$ so $f\left(x,y\right)\ge 0$ , the only point where $f\left(x,y\right)=0$ being the stationary point. This is therefore a local (and global) minimum.
Exercises
Determine the nature of the stationary points of the function in each case:
 $f\left(x,y\right)=8{x}^{2}+6{y}^{2}2{y}^{3}+5$
 $f\left(x,y\right)={x}^{3}+15{x}^{2}20{y}^{2}+10$
 $f\left(x,y\right)=4{x}^{2}xy{y}^{2}$
 $f\left(x,y\right)=2{x}^{2}+{y}^{2}+3xy3y5x+8$
 $f\left(x,y\right)={\left({x}^{2}+{y}^{2}\right)}^{2}2\left({x}^{2}{y}^{2}\right)+1$
 $f\left(x,y\right)={x}^{4}+{y}^{4}+2{x}^{2}{y}^{2}+2{x}^{2}+2{y}^{2}+1$
 $\left(0,0\right)$ local minimum, $\left(0,2\right)$ saddle point.
 $\left(0,0\right)$ saddle point, $\left(10,0\right)$ local maximum.
 $\left(0,0\right)$ local maximum.
 $\left(1,3\right)$ saddle point.
 $\left(0,0\right)$ saddle point, (1,0) local minimum, $\left(1,0\right)$ local minimum.
 $f\left(x,y\right)\equiv {\left({x}^{2}+{y}^{2}+1\right)}^{2}$ , local minimum at $\left(0,0\right)$ .