Find the absolute max / min of f(x,y)= 4xy/(x2+1)(y2+1) on the domain by D = { (x,y) : x2+y2≤1, x1, y ≥ 0}.

Question

Find the absolute max / min of f(x,y)= 4xy/(x²+1)(y²+1) on the domain by D = { (x,y) : x²+y²≤1, x, y ≥ 0}.

Answer 1

I got a derivative:

(x^2+1) y' [ (y^2 +1)(4x) - 8xy^2)] + [(x^2+1)(y^2+1)(4y) - (8x^2y)(y^2+1)] /all over/ (x^2+1)^2(y^2+1)^2

Extrema found by setting this equal to zero and solving for y':

[(8x^2y)(y^2+1)]-[(x^2+1)(y^2+1)(4y)] // all over // [ (y^2 +1)(4x) - 8xy^2)](x^2+1)

The domain is the quarter unit circle x^2 + Y^2<=1, x>=0, y>=0, which is in the first quadrant only, so the critical points are (0,0), (0,1), (1,0)

unfortunately, the derivative does not exist anywhere at these 3 points