Source: Stewart's Calculus, 8th ed., Sections 14.7, 15.1
Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, second partial derivatives, Clairaut's theorem, basic constraint equations. Read the partial derivatives study notes first.
Once you can compute partial derivatives, the natural next question is: where does a multivariable function reach its highest and lowest values? This block covers three closely related skills. First, finding and classifying local extrema using the second derivative test. Second, finding absolute extrema on closed bounded regions by checking both interior critical points and the boundary. Third, constrained optimisation, where you maximise or minimise a function subject to a constraint, often using Lagrange multipliers. These are among the most heavily tested topics on Math 126 midterms.
Set both partial derivatives to zero to find critical points, then use the discriminant D = f_xx f_yy - (f_xy)^2 to classify each one. For absolute extrema on a closed region, also check the boundary. For optimisation with a constraint, use Lagrange multipliers: set the gradient of f equal to lambda times the gradient of the constraint.
Tags: critical point, saddle point, local maximum, local minimum, second derivative test, discriminant, absolute maximum, absolute minimum, closed bounded region, Lagrange multipliers, constraint, gradient
Critical point
A point (a, b) where both f_x(a, b) = 0 and f_y(a, b) = 0, or where one of the partials does not exist.
Think of it as: a flat spot on the surface where the tangent plane is horizontal.
Saddle point
A critical point that is neither a local maximum nor a local minimum. The function increases in some directions and decreases in others.
In simple terms, it looks like a horse saddle or a Pringle crisp: curving up one way and down the other.
Discriminant (second derivative test)
D = f_xx(a, b) f_yy(a, b) - [f_xy(a, b)]^2. Used to classify critical points. D > 0 with f_xx < 0 means local max; D > 0 with f_xx > 0 means local min; D < 0 means saddle point; D = 0 is inconclusive.
Think of it as: the multivariable version of checking concavity in single-variable calculus.
Lagrange multipliers
A method for finding the extreme values of f(x, y, z) subject to a constraint g(x, y, z) = k. You solve the system grad f = lambda grad g together with g = k.
In simple terms, at an extreme point on the constraint surface, the gradient of f must point in the same direction as the gradient of the constraint.
Finding critical points
Compute f_x and f_y.
Set both equal to zero and solve the resulting system of equations.
Each solution (a, b) is a critical point.
Classifying with the discriminant
Compute f_xx, f_yy, and f_xy at the critical point.
Calculate D = f_xx f_yy - (f_xy)^2.
If D > 0 and f_xx < 0: local maximum.
If D > 0 and f_xx > 0: local minimum.
If D < 0: saddle point.
If D = 0: the test is inconclusive.
Worked pattern (from Problem 4a)
f(x, y) = xy - 2x - 2y - x^2 - y^2:
f_x = y - 2 - 2x = 0, f_y = x - 2 - 2y = 0.
Solving: x = -2, y = -2.
f_xx = -2, f_yy = -2, f_xy = 1. D = (-2)(-2) - 1 = 3 > 0 and f_xx < 0, so f(-2, -2) = 4 is a local maximum.
Multiple critical points (from Problem 4b)
f(x, y) = xy e^(-(x^2 + y^2)/2) has critical points at (0, 0), (1, 1), (-1, -1), (1, -1), (-1, 1). The origin is a saddle point; (1, 1) and (-1, -1) give local maxima of e^(-1); (1, -1) and (-1, 1) give local minima of -e^(-1). This illustrates that a single function can have several critical points of different types.
The extreme value theorem (multivariable)
If f is continuous on a closed, bounded set D, then f attains both an absolute maximum and an absolute minimum on D.
Method
Find all critical points of f inside D.
Parametrise each piece of the boundary of D (edges, curves) and find the extreme values of f on each piece.
Compare all values from the interior critical points and the boundary. The largest is the absolute maximum; the smallest is the absolute minimum.
Worked pattern (from Problem 5a)
f(x, y) = x^2 + y^2 - 2x on the triangular region with vertices (2, 0), (0, 2), (0, -2):
Interior: f_x = 2x - 2 = 0, f_y = 2y = 0, so the only interior critical point is (1, 0) with f(1, 0) = -1.
Boundary: check each of the three edges by substituting the edge equation into f and optimising a single-variable function.
Compare all candidates. The absolute maximum is f(0, 2) = f(0, -2) = 4; the absolute minimum is f(1, 0) = -1.
Circular domain (from Problem 5b)
f(x, y) = xy^2 on D = {x >= 0, y >= 0, x^2 + y^2 <= 3}:
On the circular boundary, parametrise x = sqrt(3) cos t, y = sqrt(3) sin t and optimise.
Absolute maximum is f(1, sqrt(2)) = 2; absolute minimum is 0 along the axes.
When to use Lagrange multipliers
Whenever the problem says "find the maximum/minimum of f subject to a constraint g = k." Typical setups: closest point to a surface, box dimensions with fixed volume or surface area.
Method
Write the system: f_x = lambda g_x, f_y = lambda g_y, f_z = lambda g_z (if three variables), plus the constraint g(x, y, z) = k.
Solve the system for x, y, z (and lambda, though lambda itself is usually not needed for the final answer).
Evaluate f at each solution to determine which is the max and which is the min.
Worked pattern (from Problem 6a)
Find the point on the plane x - 2y + 3z = 6 closest to (0, 1, 1). Minimise f(x, y, z) = x^2 + (y - 1)^2 + (z - 1)^2 subject to g(x, y, z) = x - 2y + 3z = 6.
grad f = (2x, 2(y-1), 2(z-1)), grad g = (1, -2, 3).
From 2x = lambda, 2(y-1) = -2 lambda, 2(z-1) = 3 lambda: x = lambda/2, y = 1 - lambda, z = 1 + 3 lambda/2.
Substitute into the constraint and solve for lambda, then back-substitute.
Result: (5/14, 2/7, 29/14).
Box optimisation (from Problem 6c)
Minimise surface area 2(xy + yz + xz) subject to volume xyz = 1000. By symmetry of the Lagrange equations, the optimal box is a cube: 10 cm on each side.
Box with fixed surface area (from Problem 6d)
Maximise volume xyz subject to 2(xy + yz + xz) = 64. Again, symmetry gives a cube with edge length 8/sqrt(6) cm.
Second derivative test discriminant
D = f_xx(a, b) f_yy(a, b) - [f_xy(a, b)]^2
Classification rules
D > 0, f_xx < 0: local maximum
D > 0, f_xx > 0: local minimum
D < 0: saddle point
D = 0: inconclusive
Lagrange multiplier system
grad f = lambda grad g, together with g(x, y, z) = k
In two variables: f_x = lambda g_x, f_y = lambda g_y, g(x, y) = k (three equations, three unknowns: x, y, lambda).
Constrained optimisation is everywhere in engineering and economics. Designing a container with minimum material for a given volume is a direct Lagrange multiplier problem. In economics, maximising utility subject to a budget constraint uses the same framework.
The second derivative test is the basis for finding optimal points in machine learning loss surfaces (though in higher dimensions the analysis is more involved).
Students often forget to check the boundary when finding absolute extrema. The absolute max or min frequently occurs on the boundary, not at an interior critical point.
A common error is setting only one partial to zero instead of both. You need f_x = 0 and f_y = 0 simultaneously.
When D = 0, students sometimes guess local max or min. The test gives no information in this case; further analysis is needed.
In Lagrange multiplier problems, students sometimes forget to include the constraint equation in their system. Without it you have more unknowns than equations and cannot solve.
Expect at least one "find and classify all critical points" problem. Have the discriminant formula memorised cold.
Absolute extrema on a triangular or circular region is a common full-length problem. Boundary work is the most time-consuming part, so practise parametrising edges.
Lagrange multiplier problems (closest point to a plane, optimal box dimensions) appear regularly. The setup is always the same: identify f and g, write the system, solve.
The problems from the practice set (4a, 4b, 4c, 5a, 5b, 5c, 6a through 6d) are representative of midterm difficulty.
True or false: If D > 0 at a critical point, the point is definitely a local extremum. (Answer: True, either a max or min depending on the sign of f_xx.)
Fill in the blank: If D < 0 at a critical point, the point is a ______. (Answer: saddle point.)
True or false: The absolute maximum of a continuous function on a closed bounded region always occurs at an interior critical point. (Answer: False. It can occur on the boundary.)
Fill in the blank: In Lagrange multipliers, you solve the system grad f = ______ grad g. (Answer: lambda.)
True or false: If D = 0, the critical point is always a saddle. (Answer: False. The test is inconclusive.)
Q: Find and classify the critical points of f(x, y) = (x^2 + y^2) e^(-x).
A: f_x = (2x - x^2 - y^2) e^(-x) = 0, f_y = 2y e^(-x) = 0. From f_y = 0, y = 0. Then 2x - x^2 = 0, so x = 0 or x = 2. Critical points: (0, 0) and (2, 0). At (0, 0), D > 0 and f_xx > 0, so local minimum f(0,0) = 0. At (2, 0), D < 0, so saddle point.
Q: Find the absolute max and min of f(x, y) = 2x^3 + y^4 on the unit disc x^2 + y^2 <= 1.
A: Interior critical point: f_x = 6x^2 = 0, f_y = 4y^3 = 0, giving (0, 0) with f = 0. On the boundary x = cos t, y = sin t, evaluate f = 2 cos^3 t + sin^4 t and find the maximum is f(1, 0) = 2 and the minimum is f(-1, 0) = -2.
Q: Find the dimensions of a rectangular box with volume 1000 cm^3 that has minimal surface area.
A: Minimise S = 2(xy + yz + xz) subject to xyz = 1000. By Lagrange multipliers, the optimal solution has x = y = z = 10 cm (a cube).
Q: Find the point on the plane x - 2y + 3z = 6 closest to (0, 1, 1).
A: Minimise (x^2 + (y-1)^2 + (z-1)^2) subject to x - 2y + 3z = 6. The closest point is (5/14, 2/7, 29/14).
The second derivative test relies on the second partial derivatives covered in the first study notes document. The discriminant D uses f_xx, f_yy and f_xy, so fluency with computing those is a prerequisite.
Lagrange multipliers connect to the gradient and level curves (Section 14.6). The geometric idea is that at a constrained extremum, the gradient of f is parallel to the gradient of g, which is why you set grad f = lambda grad g.
Absolute extrema problems on regions bounded by curves connect to the integration topics in the next study notes, since the same regions show up as domains of double integrals.
critical point multivariable, saddle point, local maximum minimum multivariable, second derivative test two variables, discriminant f_xx f_yy f_xy, absolute extrema closed region, extreme value theorem multivariable, Lagrange multipliers, constrained optimisation, closest point to plane, box optimisation surface area volume, Math 126 UW, Stewart Calculus 14.7, 15.1