This comes from my notebook and I am confused about critical points. Let me rewrite it.
Find extreme values of the function $f(x,y) = x\sqrt{y-2} + y\sqrt{x-2}$
$D_{f} = \Big((x,y) \in \mathbb{R}^2 : x \geq 2 \land y \geq 2 \Big)$
$\frac{\partial f}{\partial x} = \sqrt{y-2} + \frac{y}{2\sqrt{x-2}} \quad \quad \Longrightarrow D_{\frac{\partial f}{\partial x}} = \Big((x,y) \in \mathbb{R}^2 : x > 2 \land y \geq 2 \Big)$
$\frac{\partial f}{\partial y } = \sqrt{x-2} + \frac{x}{2\sqrt{y-2}}\quad \quad \Longrightarrow D_{\frac{\partial f}{\partial y }} = \Big((x,y) \in \mathbb{R}^2 : x \geq 2 \land y > 2 \Big)$
Solving the system of equations:
$$\begin{cases} \frac{\partial f}{\partial x } = 0 \\ \frac{\partial f}{\partial y } = 0 \\ \end{cases} \iff \begin{cases} \sqrt{y-2} + \frac{y}{2\sqrt{x-2}} = 0 \\ \sqrt{x-2} + \frac{x}{2\sqrt{y-2}} = 0 \\ \end{cases}$$
yields contradiction: no stationary points, no solutions. And normally this is where I would end up the task.
However, in my notebook suddenly this appeared:
$$\text{Critical Points:}$$
$$P(2,y) \quad y \geq 2 \quad \Longrightarrow f(2,y) = 2\sqrt{y-2} \quad \text{local min at } y = 2$$
$$P(x,2) \quad x \geq 2 \quad \Longrightarrow f(x,2) = 2\sqrt{x-2} \quad \text{local min at } x = 2$$
So the local minimum is at $P(2,2)$.
I am not sure where those critical points came from. I am certain it's related to the domain, but still not sure if it's related to domain of derivative (which one?) or domain of function…
How should I find critical points? I am looking for general advice for solving these kind of tasks, not this task specific.
I understand how the stationary points from solving the system of equations are found, I am just unsure how the magic critical points are found out of the domain.
I solved few tasks and I never had to bring up "magic solutions" from the domain even if it was weird domain.
Main problem: how can I tell the difference between considering the task finished after solving the system of linear equations, and considering the domain to find other solutions from the domain.
Best Answer
The key here is that the derivatives are only defined where $x>2, y>2$; you must check critical points on the boundary. In general, you must check all places where the derivative is not defined.