# What are the critical points, if any, of f(x,y) = 2x^3 - 6x^2 + y^3 + 3y^2 - 48x - 45y?

Mar 26, 2017

$\left.\begin{matrix}f \left(- 2 - 5\right) & = 231 & \implies \left(- 2 - 5 231\right) \\ f \left(- 2 3\right) & = - 25 & \implies \left(- 2 3 25\right) \\ f \left(4 - 5\right) & = 15 & \implies \left(4 - 5 15\right) \\ f \left(4 3\right) & = - 241 & \implies \left(4 3 - 241\right)\end{matrix}\right.$

#### 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 ${f}_{x x} , {f}_{y y}$ 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;

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

So we have:

$f \left(x , y\right) = 2 {x}^{3} - 6 {x}^{2} + {y}^{3} + 3 {y}^{2} - 48 x - 45 y$

Let us find the first partial derivatives:

$\frac{\partial f}{\partial x} = 6 {x}^{2} - 12 x - 48$

$\frac{\partial f}{\partial y} = 3 {y}^{2} + 6 y - 45$

So our critical simultaneous equations are:

$\frac{\partial f}{\partial x} = 0 \implies 6 {x}^{2} - 12 x - 48 = 0$
$\frac{\partial f}{\partial y} = 0 \implies 3 {y}^{2} + 6 y - 45 = 0$

$\therefore {x}^{2} - 2 x - 8 = 0$
$\therefore \left(x - 4\right) \left(x + 2\right) = 0$
$\therefore x = - 2 , 4$

$\therefore {y}^{2} + 2 y - 15 = 0$
$\therefore \left(y + 5\right) \left(y - 3\right) = 0$
$\therefore y = - 5 , 3$

Any permutation of these solutions will simultaneously make both partial derivatives vanish, so the critical values are:

$\left(x , y\right) = \left(- 2 , - 5\right) , \left(- 2 , 3\right) , \left(4 , - 5\right) , \left(4 , 3\right)$

So we can now calculate the coordinates of the critical points:

$\left.\begin{matrix}f \left(- 2 - 5\right) & = 231 & \implies \left(- 2 - 5 231\right) \\ f \left(- 2 3\right) & = - 25 & \implies \left(- 2 3 25\right) \\ f \left(4 - 5\right) & = 15 & \implies \left(4 - 5 15\right) \\ f \left(4 3\right) & = - 241 & \implies \left(4 3 - 241\right)\end{matrix}\right.$

We can visualise these critical points on a 3D-plot: As we were not asked to determine the nature of the critical points I will omit that analysis.