Mark M. answered 04/29/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f(x,y) = e(x^2+y^2+2x-2y) > 0 for all points (x,y) in lR2
First, find the critical points:
fx = (2x+2)e(x^2+y^2+2x-2y) = 0 when x = -1
fy = (2y-2)e(x^2+y^2+2x-2y) = 0 when y = 1
The only critical point is (-1,1)
Since fxx(x,y) = 2e(x^2+y^2+2x-2y) + (2x+2)2e(x^2+y^2+2x-2y), fxx(-1,1)) = 2e-2 > 0
Since fyy(x,y) = 2e(x^2+y^2+2x-2y) + (2y-2)2e(x^2+y^2+2x-2y) , fyy(-1,1) = 2e-2
Since fxy(x,y) = (2x+2)(2y-2)e(x^2+y^2+2x-2y), fxy(-1,1) = 0
D = fxx(-1,1)fyy(-1,1) - [fxy(-1,1)]2 = 4e-4 > 0
Since D > 0 and fxx(-1,1) > 0, there is a relative minimum at (-1,1).