Find the critical points and determine whether they are loca

Solution

df/dx=xy-1+(x-y)y=2xy-y^2-1 df/dy=-(xy-1)+(x-y)x=x^2-2xy+1 2xy-y^2-1=0 x^2-2xy+1=0 we add them x^2-y^2=0 x=y or x=-y If x=y, we replace in the above equation x^2=1, x=-1 or x=1 if x=-y, -2x^2-x^2=1 -3x^2=1 contradiction So the critical points are x=y=1 x=y=-1 The second derivatives are A=d^2f/dx^2=2y B=d^2f/dxdy=2x-2y C=d^2f/dy^2=-2x B^2-AC=4x^2+4y^2-8xy+4xy=(x+y)^2 B^2-AC>0, A>0 for (1,1) so it is a minimum A<0 for (-1,-1) so it is a maximum