# How do you find the critical points for f(x,y)=xy(1-8x-7y)?

Jun 29, 2018

There is a saddle point at $\left(0 , 0\right)$ and a local maximum at $\left(\frac{1}{24} , \frac{1}{21}\right)$

#### Explanation:

The function is

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

Caculate the partial derivatives

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

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

The critical points are

$\left\{\begin{matrix}y - 16 x y - 7 {y}^{2} = 0 \\ x - 8 {x}^{2} - 14 x y = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}y \left(1 - 16 x - 7 y\right) = 0 \\ x \left(1 - 8 x - 14 y\right) = 0\end{matrix}\right.$

Therefore, $\left(0 , 0\right)$ is a point

$\iff$, $\left\{\begin{matrix}\left(16 x + 7 y\right) = 1 \\ \left(8 x + 14 y\right) = 1\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\left(16 x + 7 y\right) = 1 \\ \left(16 x + 28 y\right) = 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}16 x + 7 y = 1 \\ y = \frac{1}{21}\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}x = \frac{1}{24} \\ y = \frac{1}{21}\end{matrix}\right.$

The other point is $\left(\frac{1}{24} , \frac{1}{21}\right)$

Calculate the second derivatives

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

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

$\frac{{\partial}^{2} f}{\partial x \partial y} = 1 - 16 x - 14 y$

$\frac{{\partial}^{2} f}{\partial y \partial x} = 1 - 16 x - 14 y$

Calculate the Determinant $D \left(x , y\right)$ of the hessian Matrix

$\left(\begin{matrix}- 16 y & 1 - 16 x - 14 y \\ 1 - 16 x - 14 y & - 14 y\end{matrix}\right)$

$D \left(x , y\right) = 224 {y}^{2} - {\left(1 - 16 x - 14 y\right)}^{2}$

Therefore,

$D \left(0 , 0\right) = - 1$

As $D \left(0 , 0\right) < 0$, this is a saddle point.

$D \left(\frac{1}{24} , \frac{1}{21}\right) = 0.51 - 0.11 = 0.4$

$D \left(\frac{1}{24} , \frac{1}{21}\right) > 0$, then $\frac{{\partial}^{2} f \left(\frac{1}{24} , \frac{1}{21}\right)}{\partial {x}^{2}} = - \frac{16}{21}$

$\frac{{\partial}^{2} f \left(\frac{1}{24} , \frac{1}{21}\right)}{\partial {x}^{2}} < 0$

This is a local maximum at $\left(\frac{1}{24} , \frac{1}{21}\right)$