What does $D = \{(x, y) : 0 ≤ x ≤ 1$ and $1 ≤ y ≤ 2\}$ look like and "repeat" extrema? I have to find the absolute min and max of $f(x, y) = x^2 - x^2y$ on D. So first my question is what does this closed, bounded set, $D = \{(x, y) : 0 ≤ x ≤ 1$    and $1 ≤ y ≤ 2\}$  look like?
I have drawn what I think it looks like, below

If my drawing is correct, To find global extrema, I am finding the max/min on the boundary and the critical points to do some comparisons. I have no problem finding the critical points, but a bit confused about the boundaries.
Can I continue this problem by investigating each of the boundary/line?
For example, setting
$x = 1, 1 ≤ y ≤ 2$ (right) ,
$x= 0, 1 ≤ y ≤ 2$ (left)
$y = 2, 0 ≤ x ≤ 1$ (top)
$y =1, 0 ≤ x ≤ 1$ (bottom)
And find the max/min on each boundary? Do i have to do this? I have done some previous questions, and realised that some of the extrema occur more than once. How can I avoid these "repeats" extrema? Do we always get "repeats"? How do you tell?
EDIT:
Also can you have critical points in general  but have NO critical points satisfying  $D = \{(x, y) : 0 ≤ x ≤ 1$    and $1 ≤ y ≤ 2\}$?
Then the max/min cannot occur here right?
 A: Since $f$ is continuous and the set $D$ is bounded and closed, by the Weierstrass theorem $f$ must attain its maximal and minimal value on $D$, but these maximum and minimum concern only the values in $D$, we do not care if the function is, say, larger somewhere outside $D$.
It may be better seen on one-dimensional examples: let $f(x) = x$ and let $D=[0,1]$. Then $f$ has no critical points whatsoever, but it attains its max and min in $D$, which are respectively, $f(1)=1$ and $f(0)=0$. Also, if we let $f(x) = (x+1)^2$, then $f$ has a critical point at $-1$, where it attains its global minimum, but the minimum of $f$ in $D$ is $f(0)=1$. Even more interestingly, $f(x)=x^3$ has a critical point at $x=0$, yet on $D=[-1,1]$ neither its maximal and minimal value is attained at this critical point.
In general, the max and min for (sufficiently regular) $f$ in $D$ may occur either at the critical points in $D$ or on the boundary (because if the extremal value was somewhere inside $D$ (i.e., not on the boundary), it would be a local extremum). Since the boundary in your example has a nice explicit description, you may (as you do) study the function restricted to the parts of the boundary – then you search for the critical points which are in the domain of the one-variable function you currently investigate. But then of course, you have to remember that the restricted functions can attain their extremal values at the ends of their respective domains, so you also have to look at the vertices of $D$.
Repeats can happen for example if the initial function has a critical point at the boundary of $D$. You could avoid them by only considering the critical points of $f$ inside $D$, but this is only a minor technicality.
To summarize, in order to answer the question you just need to collect all the points in which the max and min can be attained, that is

*

*critical points of the initial function in $D$,

*critical points of the restrictions to the parts of the boundary (edges of $D$) (but mind their respective domains!),

*points where different parts of the boundary connect (vertices of $D$),

and compare the values of $f$ at all these points. The largest is the max in $D$ and the smallest – the min in $D$.
A: You can do what you propose, that is consider critical values and consider the extremas at the boundary.
For your particular problem, we can solve it as follows:
$$f(x,y)=x^2-x^2y=x^2(1-y)\le0$$
Hence $f$ attain its maximum value when $x=0$ or when $y=1$.
For the minimum value, note that $x^2 \ge0$, we want $y$ to be as large as possible so that $1-y$ to be as negative as possible. Also, we want $x$ to be as large in magnitude as possible. Hence the minimum value occur at $(1, 2)$.

To avoid repeated solution, we can always consider cases such that each point on the boundary only occur in exactly one the case. For example, you can change the top and bottom to be :
$$=2,0<<1 \tag{top}$$
$$=1,0<<1 \tag{bottom}$$

We are only concerned about potential solution in our feasible region.
