Relative Minimums and Maximums — Question 4

Question 4 – Solution

Strategy. When the Hessian degenerates, return to the definition and compare the function along paths through the critical point.

Step 1: Critical point $$f_x=3x^2-3y^2,\qquad f_y=-6xy.$$ The second equation requires $x=0$ or $y=0$. In either case the first equation then forces the other coordinate to be zero. Thus the only critical point is $(0,0)$.

Step 2: Failed Hessian test $$f_{xx}=6x,\qquad f_{yy}=-6x,\qquad f_{xy}=-6y.$$ All three second derivatives vanish at the origin, so $$D(0,0)=0.$$ The test gives no classification.

Step 3: Direct classification Along $y=0$, $$f(x,0)=x^3,$$ which takes both signs arbitrarily close to zero. Hence the origin is a saddle.

For a structural check, let $x=r\cos\theta$ and $y=r\sin\theta$. Then $$f=r^3\bigl(\cos^3\theta-3\cos\theta\sin^2\theta\bigr) =r^3\cos(3\theta).$$ The factor $\cos(3\theta)$ alternates sign in six angular sectors. Therefore

$$\boxed{(0,0)\text{ is a degenerate saddle point}.}$$