Since ln(x) is monotonically increasing function, consider ln(f(x,y) = ylnx + xlny, for max or min, observe the partial derivatives and boundary points.
First boundaries, since they're easier: x = 0 (x actually can not take the exact value 0, but consider a small neighborhood of 0, assuming limit exists), Then f(x,y) = 1; And x = 1, then max{f(x,y)} = 2, the situation is symmetric for y = 0.
Let fx denote the partial derivative with respect to x and fy denote ... w.r.t y.
fx = y/x + lny = 0 (E1) and fy = x/y + lnx = 0 (E2) --> y/x + lny = x/y + lnx --> y/x = x/y + ln(x/y),
Take x/y = k, (k is between 0 and infinity) then we have 1/k = lnk + k, where LHS has negative derivative always, and RHS has only positive derivative in the defined domain for k. So, they must have at most one intersection point. And by inspection, that is 1.
So, x = y. Using E1 and E2, we also have lnx lny = 1 --> ln2x = 1 --> lnx = -1 (x < 1) --> x = y = 1/e.
Then max{f(x,y)} = 2/e1/e which is clearly greater than 1, but smaller than 2. Then the maximum value is 2 and minimum is 1.
