# What are the extrema and saddle points of f(x,y) = xy(e^(y^2)-e^(x^2))?

Feb 22, 2017

 {: ("Critical Point","Conclusion"), ((0,0,0),"saddle") :}

#### Explanation:

The theory to identify the extrema of $z = f \left(x , y\right)$ is:

1. Solve simultaneously the critical equations

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0 \setminus \setminus \setminus$ (ie ${f}_{x} = {f}_{y} = 0$)

2. Evaluate ${f}_{x x} , {f}_{y y} \mathmr{and} {f}_{x y} \left(= {f}_{y x}\right)$ at each of these critical points. Hence evaluate $\Delta = {f}_{x x} {f}_{y y} - {f}_{x y}^{2}$ at each of these points
3. Determine the nature of the extrema;

$\left.\begin{matrix}\Delta > 0 & \text{There is minimum if " f_(x x)<0 \\ \null & "and a maximum if " f_(yy)>0 \\ Delta<0 & "there is a saddle point" \\ Delta=0 & "Further analysis is necessary}\end{matrix}\right.$

So we have:

$f \left(x , y\right) = x y \left({e}^{{y}^{2}} - {e}^{{x}^{2}}\right)$
$\text{ } = x y {e}^{{y}^{2}} - x y {e}^{{x}^{2}}$

Let us find the first partial derivatives:

$\frac{\partial f}{\partial x} = y {e}^{{y}^{2}} + \left\{\left(- x y\right) \left(2 x {e}^{{x}^{2}}\right) + \left(- y\right) \left({e}^{{x}^{2}}\right)\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = y {e}^{{y}^{2}} - 2 {x}^{2} y {e}^{{x}^{2}} - y {e}^{{x}^{2}}$

$\frac{\partial f}{\partial y} = \left\{\left(x y\right) \left(2 y {e}^{{y}^{2}}\right) + \left(x\right) \left({e}^{{y}^{2}}\right)\right\} - x {e}^{{x}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 x {y}^{2} {e}^{{y}^{2}} + x {e}^{{y}^{2}} - x {e}^{{x}^{2}}$

So our critical equations are:

$y {e}^{{y}^{2}} - 2 {x}^{2} y {e}^{{x}^{2}} - y {e}^{{x}^{2}} = 0 \implies y \left({e}^{{y}^{2}} - 2 {x}^{2} {e}^{{x}^{2}} - {e}^{{x}^{2}}\right) = 0$
$2 x {y}^{2} {e}^{{y}^{2}} + x {e}^{{y}^{2}} - x {e}^{{x}^{2}} = 0 \implies x \left(2 {y}^{2} {e}^{{y}^{2}} + {e}^{{y}^{2}} - {e}^{{x}^{2}}\right) = 0$

From the these equations we have:

$y = 0$ or ${e}^{{y}^{2}} - {e}^{{x}^{2}} = 2 {x}^{2} {e}^{{x}^{2}}$
$x = 0$ or ${e}^{{y}^{2}} - {e}^{{x}^{2}} = - 2 {y}^{2} {e}^{{y}^{2}}$

And the only simultaneous solution is $x = y = 0$

And so we have one critical point at the origin

So, now let us look at the second partial derivatives so that we can determine the nature of the critical point (I'll just quote these results):

$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {x}^{2}} = - 4 {x}^{3} y {e}^{{x}^{2}} - 6 x y {e}^{{x}^{2}}$
$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {y}^{2}} = 4 x {y}^{3} {e}^{{y}^{2}} + 6 x y {e}^{{y}^{2}}$
$\frac{{\partial}^{2} f}{\partial x \partial y} = {e}^{{y}^{2}} - {e}^{{x}^{2}} - 2 {x}^{2} {e}^{{x}^{2}} + 2 {y}^{2} {e}^{{y}^{2}} \setminus \setminus \setminus \setminus \left(= \frac{{\partial}^{2} f}{\partial y \partial x}\right)$

And we must calculate:

$\Delta = \frac{{\partial}^{2} f}{\partial {x}^{2}} \frac{{\partial}^{2} f}{\partial {y}^{2}} - {\left(\frac{{\partial}^{2} f}{\partial x \partial y}\right)}^{2}$

at each critical point. The second partial derivative values, $\Delta$, and conclusion are as follows:

 {: ("Critical Point",(partial^2f) / (partial x^2),(partial^2f) / (partial y^2),(partial^2f) / (partial x partial y),Delta,"Conclusion"), ((0,0,0),0,0,0,= 0,"incluclusive") :}

So after all that work it is rather disappointing to get an inclusive result, but if we examine the behaviour around the critical point we can readily establish that it is a saddle point.

We can see these critical points if we look at a 3D plot: