# What are the values and types of the critical points, if any, of f(x,y)=xy(1-8x-7y)?

Feb 19, 2017

 {: ("Critical Point","Conclusion"), ((0,0,0),"saddle"), ((0,1/7,0),"saddle"), ((1/8,0,0),"saddle"), ((1/24,1/21,1/1512),"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 ${f}_{x} = {f}_{y} = 0$)

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:

$f \left(x , y\right) = x y \left(1 - 8 x - 7 y\right)$
$\text{ } = x y - 8 {x}^{2} y - 7 x {y}^{2}$

Let us find the first partial derivatives:

$\frac{\partial f}{\partial x} = y - 16 x y - 7 {y}^{2}$
$\frac{\partial f}{\partial y} = x - 8 {x}^{2} - 14 x y$

So our critical equations are:

$y - 16 x y - 7 {y}^{2} = 0$
$x - 8 {x}^{2} - 14 x y = 0$

From the First equation we have:

$y \left(1 - 16 x - 7 y\right) = 0 \implies y = 0 \mathmr{and} 1 - 16 x - 7 y = 0$

From the Second equation we have:

$x \left(1 - 8 x - 14 y\right) = 0 \implies x = 0 \mathmr{and} 1 - 8 x - 14 y$

$x = 0 \implies 1 - 7 y = 0 \implies y = \frac{1}{7}$
$y = 0 \implies 1 - 8 x = 0 \implies x = \frac{1}{8}$

And simultaneously (solution not shown) we get the solution

 x=1/24; y=1/21

And so we have four critical points with coordinates;

$\left(0 , 0 , 0\right) , \left(0 , \frac{1}{7} , 0\right) , \left(\frac{1}{8} , 0 , 0\right) , \left(\frac{1}{24} , \frac{1}{21} , \frac{1}{1512}\right)$

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

$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {x}^{2}} = - 16 y$
$\setminus \setminus \setminus \frac{{\partial}^{2} f}{\partial {y}^{2}} = - 14 x$
$\frac{{\partial}^{2} f}{\partial x \partial y} = 1 - 16 x - 14 y \setminus \setminus \setminus \setminus \left(= \frac{{\partial}^{2} f}{\partial y \partial x}\right)$

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"), ((0,0,0),0,0,1,lt 0,"saddle"), ((0,1/7,0),-16/7,0,-1,lt 0,"saddle"), ((1/8,0,0),0,-7/4,-1,lt 0,"saddle"), ((1/24,1/21,1/1512),-16/21,-7/12,-1/3,gt 0,f_(x x)<0 => "max") :}

We can see these critical points if we look at a 3D plot. The undulations in the surface are really subtle so this is a highly zoomed plot: