We are supposed to determine local extrema of:
$$ f(x,y) = \frac{1}{x} + \frac{1}{y} + xy $$
Here is what I've done so far:
- Calculate the gradient of $f$:
$$ \nabla f(x,y) = (-\frac{1}{x^2} + y, -\frac{1}{y^2} + x) $$ 2) Determine $\nabla f = \mathbf(0)$:
Neccessary optimality condition: $$ \nabla f(x_0,y_0) \overset{!}{=} \mathbf(0) \Rightarrow (x_0,y_0) = (0,0) $$
- Determine Hessian of $f$ in $(0,0)$:
$$ H = \bigg( \begin{matrix} \frac{1}{x^3} &1 \\ 1 &\frac{1}{y^3} \end{matrix} \bigg) $$
evaluated at $(0,0)$:
$$ H = \bigg( \begin{matrix} \frac{1}{0} &1 \\ 1 &\frac{1}{0} \end{matrix} \bigg) $$
where $\frac{1}{0}$ is not defined. So how can i determine the definitness of the matrix, when a major minor is not defined.
My intuition is that since non of definetness conditions are fulfilled, the Hessian is indefinite. Since it is indefinete, there is no extrema in this point.
Is my argument correct?
It does not make sense to evalute the hessian matrix in $(0,0)$, since this point it is not in the domain of your function.