Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, tangent planes, single-variable max/min (Calculus I).
This topic covers two related problems. First, finding critical points of a function of two variables and classifying them as local maxima, local minima, or saddle points using the Second Derivative Test. Second, finding the absolute (global) maximum and minimum of a function on a closed, bounded region, which requires checking both the interior critical points and all boundary edges. These problems are the multivariable analogue of the closed-interval method from Calculus I and typically carry the most points on a MATH 126 midterm.
Critical points occur where both partial derivatives are zero (or undefined). The Second Derivative Test uses f_xx, f_yy, and f_xy to classify them. For absolute extrema on a closed region, you collect all critical points inside, parametrise every boundary edge to reduce it to a single-variable problem, then compare all values to find the global max and min.
Critical point
A point (a, b) where f_x(a, b) = 0 and f_y(a, b) = 0, or where one or both partial derivatives do not exist. Think of it as a spot where the surface is momentarily flat in every direction.
Second Derivative Test (discriminant D)
A classification tool. Compute D = f_xx · f_yy - (f_xy)² at a critical point. If D > 0 and f_xx > 0, it is a local minimum. If D > 0 and f_xx < 0, it is a local maximum. If D < 0, it is a saddle point. If D = 0, the test is inconclusive.
Saddle point
A critical point that is neither a local max nor a local min. The surface curves up in one direction and down in another, like the centre of a horse saddle or a mountain pass.
Absolute (global) maximum and minimum
The largest and smallest values of f on an entire region, including its boundary. Guaranteed to exist on a closed and bounded region by the Extreme Value Theorem.
Closed bounded region
A region in the xy-plane that includes its boundary and fits inside some large circle. In simple terms, it has an edge and does not stretch to infinity.
Finding critical points
Compute f_x and f_y.
Set both equal to zero and solve the system simultaneously.
If the function involves trig (e.g. f(x, y) = x sin(y) + y²), you may get infinitely many solutions. Express them using integer parameters (e.g. y = nπ for integer n).
Worked example from the exam: f(x, y) = x sin(y) + y²
f_x = sin(y) = 0, so y = nπ for any integer n.
f_y = x cos(y) + 2y = 0.
When y = nπ: cos(nπ) = (-1)^n, so x·(-1)^n + 2nπ = 0, giving x = -2nπ·(-1)^n = 2nπ·(-1)^(n+1).
Critical points: (2nπ·(-1)^(n+1), nπ) for every integer n.
Examples: (0, 0), (2π, π), (-2π, -π), (-4π, 2π), (4π, -2π), etc.
Classifying with the Second Derivative Test
Compute f_xx, f_yy, f_xy.
Evaluate D = f_xx · f_yy - (f_xy)² at each critical point.
For f(x, y) = x sin(y) + y²: f_xx = 0, f_yy = -x sin(y) + 2, f_xy = cos(y).
At (2nπ·(-1)^(n+1), nπ): f_xx = 0, f_xy = cos(nπ) = (-1)^n, so D = 0·f_yy - ((-1)^n)² = -1.
D = -1 < 0 for all critical points except (0, 0), so all non-origin critical points are saddle points.
At (0, 0): f_xx = 0, f_yy = 2, f_xy = 1. D = 0·2 - 1² = -1 < 0. Also a saddle point.
The method (closed region method)
Find all critical points of f inside the region. Evaluate f at each.
Parametrise each boundary edge as a function of one variable. Find the critical points of that single-variable function on its interval, and evaluate f at those points and at the endpoints (vertices).
Compare every value from steps 1 and 2. The largest is the absolute max; the smallest is the absolute min.
Worked example from the exam
f(x, y) = x² + 2y² - xy - 2x - y on the triangle T with vertices (0, 0), (0, 1), (1, 1).
Step 1: Interior critical points
f_x = 2x - y - 2 = 0
f_y = 4y - x - 1 = 0
Solving: from the first equation, y = 2x - 2. Substituting into the second: 4(2x - 2) - x - 1 = 0, so 7x = 9, x = 9/7, y = 4/7.
Check: (9/7, 4/7). Is this inside T? The triangle has vertices (0,0), (0,1), (1,1). The edges are x = 0, y = 1, and y = x. For a point inside, we need 0 ≤ x ≤ 1, x ≤ y ≤ 1. Since x = 9/7 > 1, this critical point is outside T. Discard it.
Step 2: Boundary edges
Edge 1: x = 0, with 0 ≤ y ≤ 1. f(0, y) = 2y² - y. f'(y) = 4y - 1 = 0, so y = 1/4. f(0, 1/4) = 2/16 - 1/4 = -1/8.
Edge 2: y = 1, with 0 ≤ x ≤ 1. f(x, 1) = x² + 2 - x - 2x - 1 = x² - 3x + 1. f'(x) = 2x - 3 = 0, so x = 3/2, which is outside [0, 1]. No interior critical point on this edge.
Edge 3: y = x, with 0 ≤ x ≤ 1. f(x, x) = x² + 2x² - x² - 2x - x = 2x² - 3x. f'(x) = 4x - 3 = 0, so x = 3/4. f(3/4, 3/4) = 2(9/16) - 3(3/4) = 9/8 - 9/4 = -9/8.
Step 3: Evaluate at vertices
f(0, 0) = 0
f(0, 1) = 2 - 1 = 1
f(1, 1) = 1 + 2 - 1 - 2 - 1 = -1
Step 4: Compare all values
Values found: 0, -1/8, 1, -1, -9/8.
Absolute minimum = -9/8 at (3/4, 3/4). Absolute maximum = 1 at (0, 1).
Second Derivative Test discriminant
D = f_xx(a, b) · f_yy(a, b) - [f_xy(a, b)]²
D > 0 and f_xx > 0: local minimum
D > 0 and f_xx < 0: local maximum
D < 0: saddle point
D = 0: inconclusive
Critical point condition
f_x(a, b) = 0 and f_y(a, b) = 0 simultaneously.
Optimisation over bounded regions models problems like finding the temperature extremes across a metal plate, minimising cost subject to resource constraints, or locating the highest and lowest points on a terrain patch. The Second Derivative Test is the standard way to classify equilibrium points in physics and engineering stability analysis.
Students often forget to check whether the interior critical point actually lies inside the region. Solving f_x = 0, f_y = 0 gives you all critical points in the plane. You must verify each one is inside your domain.
A common error is treating the boundary as a single curve instead of breaking it into separate edges. Each straight edge of a polygon needs its own parametrisation.
Students sometimes apply the Second Derivative Test to boundary critical points. The Second Derivative Test classifies local extrema on the entire plane. On a boundary, you reduce to one variable and use single-variable methods.
When D = 0, the test says nothing. Students often assume D = 0 means saddle point. It does not. Further analysis is needed.
Absolute extrema on a closed region is one of the highest-weighted problems on a MATH 126 midterm (12 points in this exam). It tests multiple skills at once: partial derivatives, solving systems, parametrisation, and single-variable calculus.
Always list every candidate value in a final comparison table. Missing even one vertex or one boundary critical point loses marks.
For trig-based critical point problems, the exam expects you to express the full infinite family using integer parameters, not just list a few examples.
If D = 0 at a critical point, state that the test is inconclusive. Do not guess.
True or false: A critical point of f(x, y) always satisfies f_x = 0 and f_y = 0. (False. It could also be where a partial derivative is undefined.)
Fill in the blank: If D > 0 and f_xx < 0, the critical point is a local ______. (Maximum.)
True or false: The absolute maximum of a continuous function on a closed bounded region always occurs at a critical point in the interior. (False. It can occur on the boundary.)
Fill in the blank: D < 0 means the critical point is a ______ ______. (Saddle point.)
True or false: When checking a triangular boundary, you need to check exactly three edges and three vertices. (True.)
Q: Find and classify all critical points of f(x, y) = x² + y² - 4x + 6y + 13.
A: f_x = 2x - 4 = 0, so x = 2. f_y = 2y + 6 = 0, so y = -3. One critical point: (2, -3). f_xx = 2, f_yy = 2, f_xy = 0. D = 4 > 0 and f_xx > 0, so (2, -3) is a local minimum.
Q: Find the absolute max and min of f(x, y) = x + 2y on the triangle with vertices (0, 0), (3, 0), (0, 3).
A: f_x = 1, f_y = 2, so there are no critical points (f_x and f_y are never zero). Check the boundary only. On x = 0 (0 ≤ y ≤ 3): f = 2y, max at y = 3, f = 6. On y = 0 (0 ≤ x ≤ 3): f = x, max at x = 3, f = 3. On x + y = 3 (y = 3 - x, 0 ≤ x ≤ 3): f = x + 2(3 - x) = 6 - x, max at x = 0, f = 6. Vertices: f(0,0) = 0, f(3,0) = 3, f(0,3) = 6. Absolute min = 0 at (0, 0). Absolute max = 6 at (0, 3).
Q: Suppose D = 0 at a critical point. What do you conclude?
A: The Second Derivative Test is inconclusive. You would need other methods (looking at nearby values, examining the function along particular directions, or using higher-order Taylor terms) to classify the point.
Critical points and the Second Derivative Test are the multivariable generalisation of f'(x) = 0 and the second derivative test from Calculus I. The absolute extrema method extends the closed-interval method to two dimensions. This material feeds directly into Lagrange multipliers, which find extrema subject to a constraint curve rather than on the boundary of a region.
Critical points, saddle point, local max, local min, Second Derivative Test, discriminant, Hessian matrix, absolute extrema, global maximum, global minimum, closed region method, boundary parametrisation, Extreme Value Theorem, closed bounded domain, MATH 126, Calculus III, multivariable optimisation, optimization on triangles