Source: Math 126 Spring 2023 Final Exam, University of Washington
Tags: saddle points, second derivative test, tangent plane, linear approximation, absolute extrema, closed bounded region, critical points, MATH 126, Calculus III
Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, gradient (Chapter 14.1–14.4)
This is where single-variable optimisation ("find the max and min") scales up to functions of two variables. The second derivative test now involves a discriminant D = fₓₓfᵧᵧ − (fₓᵧ)², and saddle points appear as a genuinely new phenomenon with no single-variable analogue. Tangent planes generalise tangent lines and are the basis for linear approximation in higher dimensions. If you have not yet practised finding and classifying critical points, do that before reading on.
Set both partial derivatives to zero to find critical points, then use the discriminant D to classify them. For absolute extrema on a closed region, check the interior critical points and every boundary curve. Tangent planes give the best linear approximation to a surface near a point.
Critical point
A point (a, b) where both fₓ(a, b) = 0 and fᵧ(a, b) = 0 (or where a partial derivative does not exist). These are the candidates for local maxima, local minima, or saddle points.
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 middle of a horse's saddle.
Second derivative test (discriminant D)
D = fₓₓfᵧᵧ − (fₓᵧ)². If D > 0 and fₓₓ > 0, local min. If D > 0 and fₓₓ < 0, local max. If D < 0, saddle point. If D = 0, the test is inconclusive.
Tangent plane
The plane that best approximates a surface z = f(x, y) at a given point. Its equation is z = f(a, b) + fₓ(a, b)(x − a) + fᵧ(a, b)(y − b). Think of it as the flat sheet that just touches the surface at one point and tilts the same way the surface does there.
Absolute (global) extrema
The largest and smallest values a function takes on a closed, bounded region, including boundaries. The Extreme Value Theorem guarantees these exist for continuous functions on such regions.
Compute fₓ and fᵧ, set both to zero, and solve simultaneously for all critical points.
Compute fₓₓ, fᵧᵧ, and fₓᵧ. Evaluate D = fₓₓfᵧᵧ − (fₓᵧ)² at each critical point.
Where D < 0, you have a saddle point.
Example: f(x, y) = x²y² − 4x² − y².
fₓ = 2xy² − 8x = 2x(y² − 4) = 0, so x = 0 or y = ±2.
fᵧ = 2x²y − 2y = 2y(x² − 1) = 0, so y = 0 or x = ±1.
Five critical points: (0, 0), (1, 2), (−1, 2), (1, −2), (−1, −2).
fₓₓ = 2y² − 8, fᵧᵧ = 2x² − 2, fₓᵧ = 4xy.
At (0, 0): D = (−8)(−2) − 0 = 16 > 0, so (0, 0) is not a saddle point.
At (1, 2): D = (0)(0) − 8 = −8 < 0. Saddle point. Similarly for (−1, 2), (1, −2), (−1, −2).
Saddle points: (1, 2), (−1, 2), (1, −2), (−1, −2).
The tangent plane to z = f(x, y) at (a, b) is: z = f(a, b) + fₓ(a, b)(x − a) + fᵧ(a, b)(y − b).
This plane provides the linear approximation f(x, y) ≈ tangent plane value for (x, y) near (a, b).
Example: f(x, y) = x²y² − 4x² − y², at (2, 1).
f(2, 1) = 4 − 16 − 1 = −13.
fₓ(2, 1) = 2(2)(1) − 8(2) = −12.
fᵧ(2, 1) = 2(4)(1) − 2 = 6.
Tangent plane: z = −13 − 12(x − 2) + 6(y − 1).
Approximate f(2.1, 0.8): z ≈ −13 − 12(0.1) + 6(−0.2) = −15.4.
Strategy: find interior critical points, then check each boundary curve separately.
On each boundary, reduce f to a single-variable function, find its critical points, and evaluate at endpoints.
Compare all candidate values to find the absolute max and min.
Example: f(x, y) = x² − y² + 2y on the region bounded by x = 0, y = 4, and y = x².
Interior: fₓ = 2x = 0, fᵧ = −2y + 2 = 0, giving (0, 1). f(0, 1) = 1.
Top boundary (y = 4): f(x, 4) = x² − 8, critical at x = 0. Check (0, 4) and (2, 4): f = −8 and −4.
Left boundary (x = 0): f(0, y) = −y² + 2y, critical at y = 1. Check (0, 0) and (0, 4): f = 0 and −8.
Bottom-right boundary (y = x²): f(x, x²) = x² − x⁴ + 2x². f′(x) = 6x − 4x³ = 2x(3 − 2x²) = 0 at x = 0 and x = √(3/2). f(√(3/2), 3/2) = 9/4.
Absolute minimum: −8 (at (0, 4)). Absolute maximum: 9/4 (at (√(3/2), 3/2)).
D = f_{xx} f_{yy} - (f_{xy})^2\text{Tangent plane: } z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)Students often assume that D > 0 alone is enough to classify a critical point. You also need the sign of fₓₓ to distinguish between a local max and a local min.
A critical point with D = 0 is not automatically a saddle point. The test is simply inconclusive, and you would need other methods.
When finding absolute extrema, students frequently forget to parametrise and check all boundary curves. Missing one boundary means missing potential extreme values.
The tangent plane formula requires evaluating the partial derivatives at the point, not leaving them as general expressions.
⚠️ The saddle point classification via D < 0 is a near-certain exam topic. Practise the full workflow: partials to zero, solve, compute D.
⚠️ Tangent plane + linear approximation problems are typically worth 6 points and test both conceptual understanding and arithmetic.
⚠️ Absolute extrema on bounded regions are the longest problems on the exam. Budget extra time. The boundary analysis is where most students lose marks.
Fill in the blank: If D < 0, the critical point is a ______. (saddle point)
True or False: A tangent plane to z = f(x, y) at (a, b) passes through the point (a, b, f(a, b)). (True.)
True or False: To find absolute extrema on a closed region, you only need the interior critical points. (False, you must also check the boundaries.)
Q: Find all saddle points of f(x, y) = x²y² − 4x² − y².
A: Critical points are (0, 0), (1, 2), (−1, 2), (1, −2), (−1, −2). D(0, 0) = 16 > 0 (not a saddle). D < 0 at all four remaining points. Saddle points: (1, 2), (−1, 2), (1, −2), (−1, −2).
Q: Find the tangent plane to f(x, y) = x²y² − 4x² − y² at (2, 1) and approximate f(2.1, 0.8).
A: z = −13 − 12(x − 2) + 6(y − 1). f(2.1, 0.8) ≈ −15.4.
Q: Find the absolute max and min of f(x, y) = x² − y² + 2y on the region bounded by x = 0, y = 4, and y = x².
A: Absolute min = −8 at (0, 4). Absolute max = 9/4 at (√(3/2), 3/2).
The second derivative test is the multivariable extension of the concavity test from single-variable calculus. Tangent planes connect forward to directional derivatives and the gradient. Absolute extrema problems on bounded regions use the same "interior + boundary" strategy that reappears in Lagrange multipliers.
critical points multivariable, saddle point test, second derivative test two variables, discriminant D, tangent plane equation, linear approximation multivariable, absolute extrema closed region, boundary analysis, MATH 126, Calculus III, UW Math 126