Difficulty: Intermediate | Prerequisites: Critical points and the Second Derivative Test
The Second Derivative Test finds local extrema in the open interior of a surface. But when you are given a closed, bounded region (a disc, a rectangle, a triangle), the Extreme Value Theorem guarantees that absolute max and min values exist, and they can occur either in the interior or on the boundary. This section teaches the systematic method for checking both. It is one of the highest-value exam topics in MATH 126 because the problem is long, structured, and worth many marks.
To find absolute extrema on a closed bounded region: (1) find critical points in the interior and evaluate f there, (2) parametrise the boundary and find extrema of f along it, (3) compare all candidate values. The largest is the absolute max, the smallest is the absolute min.
Closed region
A region that includes its boundary. A closed disc, for example, includes all points inside the circle and all points on the circle itself. In simple terms, there is no edge you can approach without eventually reaching a point that is still in the region.
Bounded region
A region that fits inside some large circle. A disc of radius 3 is bounded; the entire xy-plane is not.
Extreme Value Theorem (for multivariable functions)
If f is continuous on a closed, bounded region D, then f attains an absolute maximum and an absolute minimum somewhere on D. Think of it as a guarantee: on a closed bounded region, the highest and lowest points on the surface must exist.
Boundary of D
The edge of the region. For a disc of radius r centred at the origin, the boundary is the circle x² + y² = r².
Parametrisation of the boundary
Rewriting the boundary curve in terms of a single variable (often an angle θ for circles, or a coordinate for line segments) so that the two-variable function becomes a single-variable function you can optimise with ordinary calculus.
Lagrange multipliers (alternative boundary method)
A technique for optimising f subject to a constraint g(x, y) = c, using the condition ∇f = λ∇g. This is an alternative to direct parametrisation for the boundary step. Think of it as finding where the gradient of f points in the same direction as the gradient of the constraint.
Step 1, interior: Find all critical points of f inside D (set f_x = 0 and f_y = 0, solve, keep only solutions that lie strictly inside D). Evaluate f at each.
Step 2, boundary: Reduce f to a single-variable function along the boundary (either by substituting the constraint directly, or by parametrising with θ for a circle). Find the critical points of that single-variable function and evaluate f at each, including the endpoints of each boundary segment.
Step 3, compare: The absolute maximum is the largest value from Steps 1 and 2. The absolute minimum is the smallest.
Step 1, interior critical points:
f_x = 2x + 4 = 0 gives x = −2. f_y = 4y = 0 gives y = 0.
The critical point is (−2, 0). Check: (−2)² + 0² = 4 ≤ 9, so it is inside D.
f(−2, 0) = 4 − 8 + 0 = −4.
Step 2, boundary (the circle x² + y² = 9):
Parametrise: x = 3 cos θ, y = 3 sin θ, with 0 ≤ θ ≤ 2π.
Substitute into f:
f = (3 cos θ)² + 4(3 cos θ) + 2(3 sin θ)² = 9 cos²θ + 12 cos θ + 18 sin²θ.
Using sin²θ = 1 − cos²θ:
f = 9 cos²θ + 12 cos θ + 18 − 18 cos²θ = −9 cos²θ + 12 cos θ + 18.
Let u = cos θ, so g(u) = −9u² + 12u + 18, with −1 ≤ u ≤ 1.
g'(u) = −18u + 12 = 0 gives u = 2/3, so cos θ = 2/3.
Evaluate g at u = 2/3, u = −1, and u = 1:
g(2/3) = −9(4/9) + 12(2/3) + 18 = −4 + 8 + 18 = 22
g(−1) = −9 + (−12) + 18 = −3
g(1) = −9 + 12 + 18 = 21
Step 3, compare all values:
Interior: f(−2, 0) = −4. Boundary: 22, −3, 21.
Absolute minimum = −4 at (−2, 0). Absolute maximum = 22 at the point(s) where cos θ = 2/3 on the circle.
For a circular boundary x² + y² = r², parametrise as:
x = r\cos\theta, \quad y = r\sin\theta, \quad 0 \le \theta \le 2\piThen substitute into f(x, y) to get a single-variable function of θ (or of u = cos θ) and use ordinary single-variable calculus.
Alternatively, use Lagrange multipliers with constraint g(x, y) = x² + y² − r² = 0:
\nabla f = \lambda \nabla gThis method is used in engineering design wherever you need to find the best performance within physical constraints, for instance the combination of dimensions that maximises the strength of a beam while keeping its cross-sectional area within a required limit. In economics, finding the production mix that maximises profit subject to a resource budget uses the same structure.
Students often check only the interior and forget the boundary entirely. On a closed region, the absolute extremum is just as likely to occur on the boundary as inside.
When parametrising a circle, students sometimes forget to check the endpoints u = −1 and u = 1 (corresponding to θ = π and θ = 0). These are critical candidates even though they are not critical points of g(u).
Mixing up "local" and "absolute" extrema. A local min in the interior might not be the absolute min if a boundary value is smaller.
Students sometimes apply the Second Derivative Test to boundary candidates. On the boundary, you are doing single-variable optimisation, not multivariable. The discriminant D is not relevant there.
⚠️ This problem type was worth 16 points on the exam, the single highest-value question. Expect it to appear on every MATH 126 midterm.
⚠️ The boundary analysis is the part students lose marks on. Practise the parametrisation step until it is automatic.
⚠️ Always present a clear comparison table at the end listing every candidate point and its f-value. Graders look for this.
⚠️ If the region is a disc, the substitution u = cos θ trick (reducing to a polynomial in u) is the fastest route.
True or False: If f has a critical point inside D, that point is automatically the absolute minimum. (False, you must compare it with boundary values.)
Fill in the blank: The Extreme Value Theorem requires D to be ____ and ____. (Closed and bounded.)
True or False: On a circular boundary, you can substitute x = r cos θ and y = r sin θ to reduce to a single-variable problem. (True.)
Fill in the blank: After finding all candidates, the absolute max is the ____ value and the absolute min is the ____ value. (Largest; smallest.)
True or False: You should use the Second Derivative Test (discriminant D) to classify boundary candidates. (False, use single-variable methods on the boundary.)
Q: Find the absolute max and min of f(x, y) = x² + y² on the closed disc x² + y² ≤ 4.
A: Interior: f_x = 2x = 0, f_y = 2y = 0 gives (0, 0) with f(0, 0) = 0. Boundary: x² + y² = 4, so f = 4 everywhere on the boundary. Absolute min = 0 at (0, 0). Absolute max = 4 on the entire circle.
Q: Why does the Extreme Value Theorem not apply on the open disc x² + y² < 4?
A: The open disc is not closed (it does not include its boundary). Without the boundary, f may approach values it never reaches, so the theorem's guarantee of attaining a max and min does not hold.
Q: In the worked example, why do you also check u = −1 and u = 1?
A: Because u = cos θ is restricted to [−1, 1], these are the endpoints of the domain for the single-variable function g(u). Endpoints are always candidates for extrema in single-variable calculus.
Q: Suppose the interior critical point lies outside D. What do you do?
A: Discard it. Only candidates inside D (or on its boundary) are relevant. The absolute extrema come entirely from the boundary analysis.
This connects to the Second Derivative Test (used for interior critical points in Step 1) and to Lagrange multipliers (an alternative method for the boundary step). It also prepares you for double integrals over bounded regions, where the same regions D reappear as domains of integration.
absolute extrema, global maximum, global minimum, Extreme Value Theorem, closed bounded region, boundary analysis, parametrisation, Lagrange multipliers, MATH 126, Calculus III, optimisation on a disc, critical points on closed region, constrained optimisation