Difficulty: Intermediate to Advanced | Prerequisites: partial derivatives, gradient, familiarity with completing the square and parametric equations.
Optimisation in several variables asks: where does a function hit its highest and lowest values, and under what conditions? On a closed, bounded region you are guaranteed an absolute max and min (the Extreme Value Theorem), but you have to check both the interior (via critical points) and every piece of the boundary. When you add a constraint like "stay on this circle," you either reduce the problem to one variable using a parametrisation or deploy Lagrange multipliers. This material ties together everything you have learnt about partial derivatives and connects directly to real applications in engineering, economics, and physics.
To find absolute extrema on a closed region, find interior critical points (set ∇f = 0), then check every boundary edge separately. Compare all candidate values. On a circle x² + y² = r², substitute x = r cos θ, y = r sin θ to reduce to a single-variable problem. The unconstrained minimum of f(x,y) = x² + y² - 2x - 4y is at (1, 2), where f = -5.
Critical point
A point where ∇f = 0 (both partial derivatives vanish) or where a partial derivative does not exist. In simple terms, it is a candidate for a local max, local min, or saddle point.
Absolute (global) maximum / minimum
The largest / smallest value f takes on the entire region, including interior and boundary. Guaranteed to exist on a closed, bounded set by the Extreme Value Theorem.
Boundary analysis
Checking each edge or curve forming the boundary of the region by restricting f to that edge (often reducing to one variable) and finding its extrema there.
Lagrange multipliers
A method for optimising f subject to a constraint g(x,y) = c. At an extremum, ∇f = λ∇g for some scalar λ. Think of it as: the gradient of f must be parallel to the gradient of the constraint.
Parametric reduction
Substituting a parametrisation of the constraint (e.g. x = r cos θ on a circle) to turn a constrained multivariable problem into a single-variable calculus problem.
The problem: f(x,y) = x² + y² - 2x - 4y on A = {(x,y) : x ≥ 0, 0 ≤ y ≤ 5, y ≥ x}
Step 1, interior critical point:
f_x = 2x - 2 = 0 → x = 1.
f_y = 2y - 4 = 0 → y = 2.
Critical point: (1, 2). Check it lies in A: x = 1 ≥ 0, y = 2, 0 ≤ 2 ≤ 5, 2 ≥ 1. Yes.
f(1, 2) = 1 + 4 - 2 - 8 = -5.
Step 2, boundary edges:
The region A is a triangle with three edges:
Edge 1: x = 0, 0 ≤ y ≤ 5. Then f = y² - 4y. f' = 2y - 4 = 0 → y = 2. f(0, 2) = -4. Endpoints: f(0, 0) = 0, f(0, 5) = 5.
Edge 2: y = 5, 0 ≤ x ≤ 5. Then f = x² - 2x + 25 - 20 = x² - 2x + 5. f' = 2x - 2 = 0 → x = 1. f(1, 5) = 4. Endpoints: f(0, 5) = 5, f(5, 5) = 20.
Edge 3: y = x, 0 ≤ x ≤ 5. Then f = 2x² - 6x. f' = 4x - 6 = 0 → x = 3/2. f(3/2, 3/2) = 2(9/4) - 9 = -9/2. Endpoints: f(0, 0) = 0, f(5, 5) = 20.
Step 3, compare all candidates:
f(1, 2) = -5 (interior)
f(0, 2) = -4 (edge 1)
f(3/2, 3/2) = -9/2 = -4.5 (edge 3)
f(0, 0) = 0, f(0, 5) = 5, f(5, 5) = 20, f(1, 5) = 4
Absolute minimum: -5 at (1, 2). Absolute maximum: 20 at (5, 5).
The unconstrained minimum is at (1, 2) with f = -5. This is the correct multiple-choice answer: (B) (1, 2).
Parametric reduction on x² + y² = 4
For f(x,y) = x² + y² - 2x - 4y on the circle x² + y² = 4:
On the circle, x² + y² = 4, so f = 4 - 2x - 4y.
Substitute x = 2 cos θ, y = 2 sin θ: f(θ) = 4 - 4 cos θ - 8 sin θ.
Differentiate: f'(θ) = 4 sin θ - 8 cos θ = 0 → tan θ = 2.
The extremal values come from evaluating f at θ = arctan 2 and θ = arctan 2 + π.
At θ = arctan 2: sin θ = 2/√5, cos θ = 1/√5, so f = 4 - 4/√5 - 16/√5 = 4 - 20/√5 = 4 - 4√5.
At θ = arctan 2 + π: f = 4 + 4√5.
Minimum on the circle: 4 - 4√5 ≈ -4.94. Maximum on the circle: 4 + 4√5 ≈ 12.94.
This parametric-reduction technique avoids Lagrange multipliers entirely and is the approach the exam hints at.
Critical point condition: f_x = 0 and f_y = 0 simultaneously.
Lagrange multipliers: ∇f = λ∇g, with g(x,y) = c as the constraint.
Parametric substitution on a circle: x = r cos θ, y = r sin θ, then optimise f(θ).
Constrained optimisation is the mathematical core of engineering design: maximise strength subject to a weight limit, minimise cost subject to a quality threshold. Economists use Lagrange multipliers to maximise utility subject to a budget constraint. The boundary-analysis technique (checking edges of a feasible region) is exactly what linear programming does.
Students often forget to check the boundary. The interior critical point may not be the absolute extremum; the boundary nearly always contributes candidates.
Confusing "critical point inside the region" with "critical point of the boundary function." These are separate calculations.
On a circle, forgetting that x² + y² is constant simplifies f before you parametrise. Use the constraint to eliminate terms.
Assuming Lagrange multipliers are always needed. On a circle, parametric reduction is often quicker and less error-prone.
⚠️ The absolute-extrema-on-a-region problem (interior + boundary) is a guaranteed exam question in most Calculus III courses.
⚠️ The unconstrained minimum of x² + y² - 2x - 4y is (1, 2). Know it cold; it is multiple choice.
⚠️ On constrained problems, the exam often says "reduce to a one-variable expression." That is code for "parametrise, do not use Lagrange multipliers."
The unconstrained minimum of x² + y² - 2x - 4y is at ___. (Answer: (1, 2), f = -5)
True or false: On a closed, bounded region, you only need to check interior critical points. (Answer: False, you must also check the boundary.)
On x² + y² = 4, f = x² + y² - 2x - 4y simplifies to f = ___. (Answer: 4 - 2x - 4y)
Q: Find the absolute max and min of f(x,y) = x² + y² - 2x - 4y on A = {x ≥ 0, 0 ≤ y ≤ 5, y ≥ x}.
A: Interior critical point (1,2) gives f = -5. Boundary analysis yields f values of -4, -4.5, 0, 4, 5, 20. Absolute min: -5 at (1,2). Absolute max: 20 at (5,5).
Q: For f(x,y) = x² + y² - 2x - 4y on x² + y² = 4, reduce to one variable and find the extremal values.
A: On the circle, f = 4 - 4 cos θ - 8 sin θ. Setting f'(θ) = 0 gives tan θ = 2. Min ≈ 4 - 4√5, max ≈ 4 + 4√5.
Q: Which point gives the unconstrained minimum of x² + y² - 2x - 4y? (A) (0,0) (B) (1,2) (C) (2,1) (D) (5,5)
A: (B) (1, 2).
This material connects to Lagrange multipliers (the formal constrained-optimisation tool), second-derivative tests for classifying critical points, and the Extreme Value Theorem from single-variable calculus. Later in the course, optimisation appears again when you maximise flux or minimise surface area integrals.
absolute extrema, global max, global min, critical point, closed bounded region, boundary analysis, constrained optimisation, Lagrange multipliers, parametric reduction, circle constraint, Extreme Value Theorem, MATH 215, Calculus III, University of Michigan