We determine all the stationary points of z(x,y)=5x−x2−y2+xy.
- z′x(x,y)=5−2x+y,
- z′y(x,y)=−2y+x.
z′y(x,y)=0 gives x=2y. We plug this into z′(x,y)=0, which gives 5−2(2y)+y=5−3y=0. Hence, y=53 and x=2y=253=103.
Hence, the stationary point is (x,y)=(103,53).