Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, gradient, dot product, parametric curves
Optimisation in several variables is where all the machinery of partial derivatives, gradients, and second-order tests comes together. You need to find where functions reach their extreme values, both on open domains (critical points, the second derivative test) and on constrained domains (Lagrange multipliers). Understanding the geometry of quadratic surfaces helps you classify what kind of extremum you have found. Particle motion ties in through velocity, speed, and arc length, all computed from derivatives of vector-valued functions.
Lagrange multipliers find extrema of f subject to a constraint g = 0 by solving ∇f = λ∇g. A zero gradient does not guarantee a local max or min; it could be a saddle point. The second derivative test (using the discriminant D = f_xx f_yy − f_xy²) classifies critical points. Speed is the magnitude of the velocity vector, and constant speed implies velocity is perpendicular to acceleration.
Critical point
A point where ∇f = 0 (all partial derivatives vanish). This is a necessary condition for a local max or min, but not sufficient: the point could also be a saddle point. Think of it as a place where the surface is "flat" in every direction.
Saddle point
A critical point that is neither a local max nor a local min. The function increases in some directions and decreases in others. The classic example is f(x, y) = x² − y² at the origin. In simple terms, it looks like the centre of a horse saddle.
Lagrange multiplier (λ)
A scalar introduced when optimising f(x, y) subject to a constraint g(x, y) = 0. At the optimum, ∇f = λ∇g, meaning the gradient of f is parallel to the gradient of the constraint. Think of it as the "exchange rate" between the objective function and the constraint.
Hyperbolic paraboloid
A saddle-shaped surface. The quadratic form z = y² − 4xy has a discriminant B² − 4AC > 0 (with A = 0, B = −4, C = 1), confirming a saddle shape.
Speed of a particle
The magnitude of the velocity vector: |r'(t)|. Speed is always non-negative and is a scalar quantity.
Arc length
The total distance a particle travels along its path from t = a to t = b, computed as the integral of speed: L = ∫ from a to b of |r'(t)| dt.
To optimise f(x, y) subject to g(x, y) = 0, solve the system: ∇f = λ∇g and g(x, y) = 0.
This gives a system of equations: fₓ = λgₓ, f_y = λg_y, and the constraint equation.
Worked example: f(x, y) = xy, constraint x² + 2y² = 1.
∇f = ⟨y, x⟩, ∇g = ⟨2x, 4y⟩.
System: y = 2λx, x = 4λy.
From the first equation, λ = y/(2x). Substituting into the second: x = 4y²/(2x), so x² = 2y².
Substituting into the constraint: 2y² + 2y² = 1, giving y = ±1/2 and x = ±1/√2.
Maximum value: 1/(2√2). Minimum value: −1/(2√2).
A critical point has ∇f = 0.
The second derivative test uses the discriminant D = f_xx f_yy − (f_xy)².
D > 0 and f_xx > 0: local minimum.
D > 0 and f_xx < 0: local maximum.
D < 0: saddle point.
D = 0: test is inconclusive.
A zero gradient does not guarantee a max or min. Example: f(x, y) = x² − y² has ∇f = 0 at the origin, but the origin is a saddle point.
For z = Ax² + Bxy + Cy², classify using the discriminant B² − 4AC.
B² − 4AC < 0: elliptic paraboloid (bowl shape).
B² − 4AC > 0: hyperbolic paraboloid (saddle shape).
B² − 4AC = 0: degenerate (parabolic cylinder).
Example: z = y² − 4xy has A = 0, B = −4, C = 1. Discriminant: 16 − 0 = 16 > 0, so it is a hyperbolic paraboloid.
Two functions can each have a saddle point at the origin, but their sum need not.
Example: f = x² − y² and g = y² − x² both have saddle points at (0, 0), but f + g = 0, which is a flat plane with no saddle point.
Given r(t) = ⟨2(t − sin t), 2 cos t, 3 cos t⟩:
Velocity: r'(t) = ⟨2(1 − cos t), −2 sin t, −3 sin t⟩.
Speed: |r'(t)| = √(4(1 − cos t)² + 4 sin² t + 9 sin² t) = √(17 − 8 cos t − 9 cos² t).
Arc length from 0 to π: L = ∫₀^π |r'(t)| dt.
If a particle moves with constant speed, then |v|² = v · v is constant.
Differentiating: 2v · a = 0, so v and a are perpendicular.
Constant speed does not mean zero acceleration (the particle can still turn), and it does not mean the path is a straight line.
\nabla f = \lambda \nabla g \quad \text{(Lagrange condition)}D = f_{xx} f_{yy} - (f_{xy})^2 \quad \text{(second derivative test discriminant)}|\mathbf{r}'(t)| = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \quad \text{(speed)}L = \int_a^b |\mathbf{r}'(t)|\, dt \quad \text{(arc length)}Lagrange multipliers are used in economics to maximise utility subject to a budget constraint, and in engineering to optimise designs subject to physical constraints (weight limits, material properties). Arc length calculations appear in robotics (how far does the end of a robot arm travel?) and in GPS trajectory analysis.
Students often assume that ∇f = 0 guarantees a local max or min. It does not; saddle points also satisfy this condition. You must use the second derivative test to classify.
Students sometimes think constant speed means zero acceleration. Constant speed means the acceleration is perpendicular to the velocity (the particle is turning, not speeding up or slowing down).
Students sometimes assume that if two functions each have a saddle point at the origin, their sum must as well. The counterexample f = x² − y² and g = y² − x² shows this is false.
When classifying z = y² − 4xy, students sometimes forget to check the discriminant B² − 4AC and instead try to "complete the square" incorrectly.
⚠️ Lagrange multipliers: set up the system ∇f = λ∇g carefully. A common error is forgetting the constraint equation as the third equation.
⚠️ "∇f = 0 implies local max or min" is a classic true/false trap. The answer is false (saddle points).
⚠️ Constant speed questions test whether you know that v · a = 0, not that a = 0.
⚠️ Surface classification: know how to compute B² − 4AC for a quadratic form and what each outcome means.
True or false: ∇f = 0 at (x₀, y₀) guarantees a local max or min. (False; it could be a saddle point)
Fill in the blank: at a constrained optimum, ∇f is ______ to ∇g. (parallel)
True or false: if a particle moves at constant speed, its acceleration must be zero. (False; acceleration is perpendicular to velocity)
Fill in the blank: z = y² − 4xy describes a ______ paraboloid. (hyperbolic)
True or false: if f and g each have a saddle point at (0, 0), then f + g must also have a saddle point at (0, 0). (False)
Q: Use Lagrange multipliers to find the max and min of f(x, y) = xy subject to x² + 2y² = 1.
A: Set up ∇f = λ∇g: y = 2λx, x = 4λy. Eliminate λ to find x² = 2y². Substitute into the constraint: 4y² = 1, so y = ±1/2 and x = ±1/√2. Max = 1/(2√2), Min = −1/(2√2).
Q: What surface does z = y² − 4xy describe?
A: A hyperbolic paraboloid (saddle shape). The discriminant B² − 4AC = 16 > 0.
Q: If a particle moves with constant speed, which must be true: (a) acceleration is zero, (b) velocity is constant, (c) velocity and acceleration are perpendicular, or (d) the path is a straight line?
A: (c). Constant speed means |v|² = v · v is constant; differentiating gives v · a = 0.
Q: True or false: ∇f(x₀, y₀) = 0 means f has a local max or local min at (x₀, y₀).
A: False. It could be a saddle point, as with f(x, y) = x² − y² at the origin.
Lagrange multipliers rely on the gradient being parallel to the constraint gradient, so a solid understanding of gradients and level sets is a prerequisite. The second derivative test is the multivariable extension of the concavity test from single-variable calculus. Arc length integrates speed, connecting parametric curves to integration techniques covered in the multiple integrals topic.
Lagrange multiplier, constrained optimisation, critical point, saddle point, second derivative test, discriminant, Hessian, hyperbolic paraboloid, elliptic paraboloid, quadratic form, surface classification, speed, velocity, acceleration, arc length, constant speed, perpendicular acceleration, parametric curve, vector-valued function, MATH 215, Calculus III, multivariable calculus, University of Michigan