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?
Best Answer
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 :
$$𝑦=2,0<𝑥<1 \tag{top}$$
$$𝑦=1,0<𝑥<1 \tag{bottom}$$
We are only concerned about potential solution in our feasible region.