Difficulty: Intermediate | Prerequisites: Partial derivatives, second-order partials (§14.3–14.4)
Big picture
Finding and classifying critical points is the multivariable version of the first- and second-derivative tests from single-variable calculus. You set both partial derivatives to zero, then use a discriminant built from the second-order partials to decide whether you are sitting on a hilltop, a valley, or a saddle. This is the foundation for unconstrained optimisation, and the next step toward Lagrange multipliers.
TL;DR
Set f_x = 0 and f_y = 0 simultaneously to find critical points. Classify each one using the second derivative test: compute D = f_xx f_yy – (f_xy)². Positive D with f_xx > 0 means a local minimum; positive D with f_xx < 0 means a local maximum; negative D means a saddle. D = 0 is inconclusive and requires other methods.
Critical point (stationary point)
A point (a, b) where f_x(a, b) = 0 and f_y(a, b) = 0, or where one or both partial derivatives fail to exist.
Think of it as a flat spot on the surface: the tangent plane is perfectly horizontal.
Second derivative test (multivariable)
A classification tool that uses the Hessian determinant D = f_xx f_yy – (f_xy)² at a critical point to determine its nature.
In simple terms, D tells you whether the surface curves the same way in all directions (extremum) or curves up in one direction and down in another (saddle).
Discriminant D (Hessian determinant)
D = f_xx f_yy – (f_xy)². Positive D means the surface curves consistently; negative D means it does not.
Saddle point
A critical point where D < 0. The function increases in some directions and decreases in others, like the centre of a horse saddle or a mountain pass.
Local maximum / local minimum
A critical point where D > 0 and f_xx < 0 (local max) or f_xx > 0 (local min). The surface curves downward or upward, respectively, in every direction from that point.
Find critical points
f_x = 3x² – 3 + 3y² = 3(x² + y² – 1) = 0
f_y = 6xy = 0
From f_y = 0: either x = 0 or y = 0.
If x = 0: f_x gives y² = 1, so y = ±1. Critical points: (0, 1) and (0, –1).
If y = 0: f_x gives x² = 1, so x = ±1. Critical points: (1, 0) and (–1, 0).
Classify using the second derivative test
f_xx = 6x, f_yy = 6x, f_xy = 6y
D = (6x)(6x) – (6y)² = 36x² – 36y² = 36(x² – y²)
At (1, 0): D = 36 > 0, f_xx = 6 > 0. Local minimum.
At (–1, 0): D = 36 > 0, f_xx = –6 < 0. Local maximum.
At (0, 1): D = –36 < 0. Saddle point.
At (0, –1): D = –36 < 0. Saddle point.
Find critical points
f_x = y – 2xy – y² = y(1 – 2x – y) = 0
f_y = x – x² – 2xy = x(1 – x – 2y) = 0
From f_x = 0: y = 0 or y = 1 – 2x.
From f_y = 0: x = 0 or x = 1 – 2y.
Case 1: x = 0 and y = 0. Critical point (0, 0).
Case 2: x = 0 and y = 1 – 2(0) = 1, but check f_y: 0(1 – 0 – 2) = 0. Critical point (0, 1). But f_x at (0,1): 1(1–0–1) = 0. Yes, this works. However f at (0,1) = 0. Actually let's be more systematic.
Case y = 0: f_y = x(1 – x) = 0, so x = 0 or x = 1. Critical points: (0, 0) and (1, 0).
Case y = 1 – 2x (and y ≠ 0): f_y = x(1 – x – 2(1–2x)) = x(1 – x – 2 + 4x) = x(3x – 1) = 0. So x = 0 or x = 1/3.
If x = 0: y = 1. Check: f_x = 1(1–0–1) = 0, f_y = 0. Critical point (0, 1).
If x = 1/3: y = 1 – 2/3 = 1/3. Critical point (1/3, 1/3).
By symmetry (swapping the roles of x and y in the case analysis), also (1, 0) (already found) and checking x = 0, y = 0 case gives us the full list: (0, 0), (1, 0), (0, 1), (1/3, 1/3).
Classify
f_xx = –2y, f_yy = –2x, f_xy = 1 – 2x – 2y
D = (–2y)(–2x) – (1–2x–2y)² = 4xy – (1–2x–2y)²
At (0, 0): D = 0 – 1 = –1 < 0. Saddle.
At (1, 0): D = 0 – (–1)² = –1 < 0. Saddle.
At (0, 1): D = 0 – (–1)² = –1 < 0. Saddle.
At (1/3, 1/3): D = 4/9 – (1 – 2/3 – 2/3)² = 4/9 – 1/9 = 3/9 = 1/3 > 0. f_xx = –2/3 < 0. Local maximum.
Expand first
f = x – xy – y + xy² = x – x²y – y + xy²
Wait, let us expand carefully: (x – y)(1 – xy) = x – x²y – y + xy².
Find critical points
f_x = 1 – 2xy + y² = 0
f_y = –1 – x² + 2xy = 0
Add the two equations: (1 – 2xy + y²) + (–1 – x² + 2xy) = y² – x² = 0, so y = ±x.
Case y = x: f_x = 1 – 2x² + x² = 1 – x² = 0, so x = ±1. Critical points: (1, 1) and (–1, –1).
Case y = –x: f_x = 1 + 2x² + x² = 1 + 3x² = 0. No real solutions.
So the critical points are (1, 1) and (–1, –1).
Classify
f_xx = –2y, f_yy = 2x, f_xy = –2x + 2y
At (1, 1): f_xx = –2, f_yy = 2, f_xy = 0. D = (–2)(2) – 0 = –4 < 0. Saddle.
At (–1, –1): f_xx = 2, f_yy = –2, f_xy = 0. D = (2)(–2) – 0 = –4 < 0. Saddle.
Both critical points are saddles.
Find critical points
f_x = 2x – 4y = 0, so x = 2y
f_y = 8y – 4x = 0, so 8y – 8y = 0, which is always true when x = 2y
Every point of the form (2t, t) is a critical point. This is a line of critical points, not an isolated one.
Classify
f_xx = 2, f_yy = 8, f_xy = –4
D = (2)(8) – (–4)² = 16 – 16 = 0
The second derivative test is inconclusive (D = 0 everywhere along the critical line).
Alternative approach
Rewrite f: x² + 4y² – 4xy + 2 = (x – 2y)² + 2.
This is a perfect square plus 2, so f ≥ 2 everywhere, with equality when x = 2y. Every point on the line x = 2y is a global minimum, with minimum value 2. There are no local maxima or saddles.
D = f_{xx} f_{yy} - (f_{xy})^2Classification at a critical point where f_x = f_y = 0:
D > 0 and f_xx > 0: local minimum
D > 0 and f_xx < 0: local maximum
D < 0: saddle point
D = 0: inconclusive (try completing the square, examining level curves, or checking values along paths)
Students often forget that D = 0 does not mean "neither max nor min." It means the test gives no answer; you need another method. Completing the square is the go-to fallback.
A line of critical points (as in Q2(d)) is unusual but not rare on exams. If your system f_x = 0, f_y = 0 gives a family of solutions rather than isolated points, recognise that D will be zero and switch strategies.
Students sometimes check only f_xx to classify, skipping the D calculation. f_xx > 0 alone does not guarantee a local minimum; you need D > 0 as well.
When factoring f_x and f_y, be systematic about case analysis. Missing a case (such as y = 0 when the factor is y(…) = 0) means missing a critical point.
⚠️ Expect at least one problem where D = 0 and you must use an algebraic trick (completing the square, examining f along lines through the critical point).
⚠️ The second derivative test formula D = f_xx f_yy – (f_xy)² is the single most-tested formula in this unit. Know it cold.
True or False: If D > 0, the critical point is definitely an extremum. (Answer: True)
Fill in the blank: If D < 0, the critical point is a ______. (Answer: saddle point)
True or False: f(x, y) = x² + y² has a saddle at the origin. (Answer: False, it has a local minimum)
Q: Find and classify the critical points of f(x, y) = x² + y² – 2x – 4y + 5.
A: f_x = 2x – 2 = 0 gives x = 1. f_y = 2y – 4 = 0 gives y = 2. One critical point: (1, 2). f_xx = 2, f_yy = 2, f_xy = 0, D = 4 > 0, f_xx > 0. Local minimum.
Q: Why can the second derivative test fail (D = 0)?
A: D = 0 means the second-order information is balanced on a knife edge: the surface is so flat at that point that second derivatives alone cannot distinguish a bowl, a hilltop, or a saddle. Higher-order behaviour or algebraic rewriting decides it.
Q: For f(x, y) = (x – 2y)² + 3, what are the critical points and their classification?
A: f_x = 2(x–2y) = 0, f_y = –4(x–2y) = 0. Both give x = 2y. The entire line x = 2y consists of critical points, all global minima with f = 3. D = 0 everywhere, but the completed-square form shows f ≥ 3.
Critical-point classification is the unconstrained optimisation prerequisite for Lagrange multipliers (§14.8), which handle constrained optimisation. The second derivative test here becomes the bordered Hessian test in the constrained case. The algebraic technique of completing the square (used in Q2(d)) reappears in linear algebra when diagonalising quadratic forms.
Related Terms / Search Tags
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