Relative Minimums and Maximums — Question 5

Question 5 – Solution

Strategy. Set $s=x^2+y^2$ and analyze the one-variable radial profile $g(s)=se^{-s}$.

Step 1: Critical set Since $$f_x=2xe^{-s}(1-s),\qquad f_y=2ye^{-s}(1-s),$$ both derivatives vanish when $(x,y)=(0,0)$ or when $s=1$. Thus $$\boxed{\text{critical set}=\{(0,0)\}\cup\{x^2+y^2=1\}}.$$

Step 2: Radial classification For $s\ge 0$, $$g'(s)=e^{-s}(1-s).$$ It is positive for $0\le s<1$ and negative for $s>1$. Thus $g(0)=0$ increases immediately away from the origin, so $$\boxed{(0,0)\text{ is a strict relative minimum}.}$$ The radial profile peaks at $s=1$, so every point of the unit circle is a relative maximum with value $$\boxed{e^{-1}}.$$

Step 3: Non-strictness and degeneracy Moving tangentially along the circle leaves $s$ and $f$ unchanged, so none of these maxima is strict. That zero-curvature tangential direction makes the Hessian determinant zero on the critical circle, which is why the standard second derivative test is inconclusive there.