You find a critical point of f(x,y) where D = f_xx · f_yy − (f_xy)² = 25 and f_xx = −5. What type of critical point is this?
AA saddle point, because f_xx is negative
BA local minimum, because D > 0
CA local maximum, because D > 0 and f_xx < 0
DAn inflection point, because D is positive but one second derivative is negative
When D > 0 and f_xx < 0, the Hessian is negative definite: the function curves downward in all directions at this point, making it a local maximum. When D > 0 and f_xx > 0, it is a local minimum. When D < 0, it is a saddle point. The sign of f_xx alone does not determine the type — both D and f_xx must be considered together.
Question 2 Multiple Choice
A student finds all critical points of f(x,y) on the closed bounded domain {(x,y) : x² + y² ≤ 4}, classifies them using the Hessian test, and identifies a local minimum in the interior. She concludes this is the global minimum. What has she most likely forgotten?
AShe forgot to verify that D > 0 at the critical point
BShe forgot to check the boundary of the domain, where the global minimum might be located
CShe forgot to compute f_xy and verify it equals f_yx
DShe forgot that the Hessian test only applies to unbounded domains
On a closed bounded domain, global extrema can occur either at interior critical points or on the boundary. A local interior minimum is only guaranteed to be global if you have also checked boundary behavior and found no smaller values there. For a disc, this means parameterizing the boundary circle and optimizing f along it. Forgetting this step is described as 'the most common source of wrong answers in applied optimization problems.'
Question 3 True / False
A critical point where D = f_xx · f_yy − (f_xy)² < 0 is a saddle point — a local minimum in one direction and a local maximum in a perpendicular direction.
TTrue
FFalse
Answer: True
When D < 0, the Hessian is indefinite — it has both positive and negative eigenvalues. The function curves upward in some directions and downward in others from that point. Like a mountain pass, the point is a minimum if you walk along the ridge but a maximum if you walk across it. This is neither a local max nor a local min overall.
Question 4 True / False
Setting ∇f = 0 at a point is sufficient to conclude that the point is a local maximum or minimum of f.
TTrue
FFalse
Answer: False
∇f = 0 is necessary but not sufficient for an extremum. It identifies all candidates — called critical points — but some are saddle points (D < 0), not extrema. If the second-derivative test is inconclusive (D = 0), further analysis is needed. Setting ∇f = 0 is the first step; classification requires the Hessian test; global optimality also requires checking boundary behavior.
Question 5 Short Answer
Explain the role of D = f_xx · f_yy − (f_xy)² in the second-derivative test. What does its sign tell you, and why can't you conclude anything from f_xx alone?
Think about your answer, then reveal below.
Model answer: D is the determinant of the Hessian matrix, which encodes curvature in all directions. D > 0 means the Hessian is definite (consistent curvature — either all upward or all downward), so the critical point is an extremum. D < 0 means the Hessian is indefinite (mixed curvature) — a saddle point. f_xx alone tells you curvature only in the x-direction; the cross terms f_xy capture how curvature in one direction depends on the other, and D combines the full picture.
The intuition: f_xx could be positive (bowl-shaped in x) while f_yy is negative (inverted in y) — that's a saddle. The determinant D = f_xx·f_yy − f_xy² is the product of the Hessian's eigenvalues, so its sign tells you whether both eigenvalues share a sign (definite — extremum) or have opposite signs (indefinite — saddle).