Finding extreme values from the domain itself

Question

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.

Answer 1

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.

Answer 2

When finding extreme values, you need to consider the boundary of the domain separately. In this case, $\partial D_f = \{(x,y) \in \mathbb{R}^2 \mid x = 2, y \ge 2\} \cup \{(x,y) \in \mathbb{R}^2 \mid x \ge 2, y = 2\}$. Thus setting $x = 2$, one finds that $y = 2$ is a minimum along the boundary. Do the same for $y = 2$. Since no other critical points exist, you conclude that this is the minimum across the entire domain.