Identify and determine the nature of the critical points of

Solution

Calculate the partial derivatives fx=2x-2xy fy=-3y^2-x^2+1 fx=0 ==> 2x(1-y)=0 ==> x=0 or y=1 fy=0 ==> if x=0, y=+-sqrt(1/3) and if y=1, x=+-sqrt(2) Look at the second partial derivatives (Hessian matrix) fxx=2-2y fxy=fyx=-2x fyy=-6y At (0,sqrt(1/3)) fxx=2-2/sqrt(3)>0, fxy=0, fyy=-6/sqrt(3)=-2sqrt(3)<0 So, the determinant is negative and this is a saddle point At (0,-sqrt(1/3)) fxx=2+2/sqrt(3)>0, fxy=0, fyy=6/sqrt(3) > 0 So, the determinant is positive and fxx is positive, so the point is a local minima At (sqrt(2),1) fxx=0, fxy=-2sqrt(2), fyy=-6 The determinant is 0-(-2sqrt(2))^2<0 and the point is a saddle point At (-sqrt(2),1) fxx=0, fxy=2sqrt(2), fyy=-6 The determinant is again 0-(2sqrt(2))^2<0 and the point is a saddle point To sum it: local minima at (0,-sqrt(3)/3) saddle points at (0,sqrt(3)/3), (sqrt(2),1) and (-sqrt(2), 1)