Find the absolute maximum and minimum values of f on the set d. f(x, y) = x2 + y2 + x2y + 6, d = {(x, y) | |x| ≤ 1, |y| ≤ 1}

The set is the square with side length 2 centered at the origin, with vertices at .

First check for possible critical points, where the first-order partial derivatives vanish:




If , then ; if , then . However, , so any points with this -coordinate fall outside of . Thus only one critical point occurs at (0, 0) within the given region.

Check the Hessian of at this point:




This matrix is positive definite, which means attains a minimum at (0, 0) with a value of .

Now we consider the boundary:

If , then , which attains a max when and a min when . At these points, we have and .

If , then and we get the same critical points as in the previous case.

If , then , which attains a max when and a min when . At these values, we get and .

If , then .

So we found an absolute maximum of 9 at (-1, 1) and (1, 1), and an absolute minimum of 6 at (0, 0).

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.