Steps to calculate a second order partial derivative

Question

Given that $$f(x, y) = x^3 − 12xy + 8y^3$$ I'm going through a series of steps to find a saddle point.

From what I've calculated:

$$f_x(x, y) = 3x^2-12y$$

$$f_y(x,y) = 24y^2-12x$$

Thus, if $f_{x}$ = 0, that means $x^2 = 4y$; And if $f_y$ = 0, that means $x = 2y^2$

Substituting $x = 2y^2$ into $x^2 = 4y$ allows us to solve: $$4y^4 − 4y = 0$$ and obtain $y = 0$ and $y = 1$. If y = 0, then x = 0, and if y = 1, then x = 2

The part I'm stuck is calculating $D(0, 0)$ where: $$D(x,y) = f_{xx}f_{yy} - (f_{xy})^2$$

I'm slightly lost after a few attempts to try and calculate $D(0, 0)$ and $D(2, 1)$ in my effort to find the saddle point for the function. How would I continue from here?

Answer 1

HINT:

$$ f(x,y) = x^3 - 12xy + 8y^3$$

The critical points are: $(0,0)$ and $(2,1)$. The partial derivatives are:

• $f_x = 3x^2 -12 y$
• $f_y = -12x + 24y^2$
• $f_{xx} = 6x$
• $f_{yy} = 48y$
• $f_{xy} = f_{yx} = -12$

The Hessian determinant is given by:

$$\det(H) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{vmatrix} \ $$

If $\det(H) \lt 0 $ then there is saddle point. For both the points $(0,0)$ and $(2,1)$, the Hessian determinant does not satisfy the condition $\det(H) \lt 0 $, so none of it is a saddle point!

Answer 2

$f_{xy} = \frac{\partial}{\partial x} (\frac{\partial}{\partial y}f(x,y))$

$\implies f_{xy} = \frac{\partial}{\partial x}f_{y} = \frac{\partial}{\partial x}(24y^2-12x) = -12$