Find the highest point on the curve of intersection of the s

Solution

g=x^2+y^2-z^2 h=x+2z-1 we need to maximize f=z grad f=s grad g+t grad h 0=2xs+t 0=2ys 1=-2zs+2t x^2+y^2-z^2=0 x+2z-1=0 so s=0 or y=0 Case 1 if s=0, then t=0 contradiction with the third equation (we get 1=0) Case 2 y=0 we get from the fourth equation x^2=z^2 so either x=z or x=-z Case 2.1 if x=z the first three equations are consistent and from the fifth equation we get 3x=1, x=1/3, so z=1/3 if x=-z the first three equations are consistent and from the fifth equation we get -x=1, x=-1, so z=1 So the maximum is 1 at x=-1,y=0,z=1