What are the extrema and saddle points of f(x,y) = 2x^3 + xy^2 + 5x^2 + y^2?

Feb 17, 2017

 {: ("Critical Point","Conclusion"), ((0,0), "min"), ((-1,-2), "saddle"), ((-1,2), "saddle"), ((-5/3,0),"max") :}

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 ${z}_{x} = {z}_{y} = 0$)

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:

$f \left(x , y\right) = 2 {x}^{3} + x {y}^{2} + 5 {x}^{2} + {y}^{2}$

Let us find the first partial derivatives:

$\frac{\partial f}{\partial x} = 6 {x}^{2} + {y}^{2} + 10 x$
$\frac{\partial f}{\partial y} = 2 x y + 2 y$

So our critical equations are:

$6 {x}^{2} + {y}^{2} + 10 x = 0$
$2 x y + 2 y = 0$

From the second equation we have:

$2 y \left(x + 1\right) = 0 \implies x = - 1 , y = 0$

Subs $x = - 1$ into the First equation and we get:

$6 + {y}^{2} - 10 = 0 \implies {y}^{2} = 4 \implies y = \pm 2$

Subs $y = 0$ into the First equation and we get:

$6 {x}^{2} + {0}^{2} + 10 x = 0 \implies 2 x \left(3 x + 5\right) = 0 \implies x = - \frac{5}{3} , 0$

And so we have four critical points with coordinates;

$\left(- 1 , - 2\right) , \left(- 1 , 2\right) , \left(0 , 0\right) , \left(- \frac{5}{3} , 0\right)$

So, now let us look at the second partial derivatives so that we can determine the nature of the critical points:

$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {x}^{2}} = 12 x + 10$
$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {y}^{2}} = 2 x + 2$
$\frac{{\partial}^{2} f}{\partial x \partial y} = 2 y \setminus \setminus \setminus \setminus \left(= \frac{{\partial}^{2} f}{\partial y \partial x}\right)$

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),10,2,0,gt 0,f_(x x)>0 => "min"), ((-1,-2),-2,0,4,lt 0,"saddle"), ((-1,2),-2,0,4,lt 0,"saddle"), ((-5/3,0),-10,-4/3,0,gt 0,f_(x x)<0 => "max") :}

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