Difficulty: Intermediate | Prerequisites: Partial derivatives, critical points, second derivative test
Once you can find and classify critical points, the next question is: what happens when the domain is not all of R²? On a bounded region (a closed shape in the xy-plane), the extreme value theorem guarantees that a continuous function has an absolute max and an absolute min. But those extremes might live on the boundary, not at an interior critical point. This section also covers minimum distance problems, a classic optimisation application where you minimise a squared-distance function subject to a surface constraint.
To find absolute extrema on a bounded domain, find all interior critical points and all critical points along the boundary, then compare f at every candidate. The biggest value is the absolute max, the smallest is the absolute min. Minimum distance problems rewrite the distance from a point to a surface as an optimisation problem on two variables.
Absolute maximum (global maximum)
The largest value of f(x, y) over the entire domain, including the boundary. Unlike a local max, there is no "nearby" qualification; it is the biggest value, full stop.
Absolute minimum (global minimum)
The smallest value of f(x, y) over the entire domain, including the boundary.
Bounded domain
A region in the xy-plane that is enclosed (has a boundary you can draw) and does not extend to infinity. Think of it as the fence around a field: you are only looking at values inside and on the fence.
Boundary
The edge of the domain. On a bounded domain, the boundary is typically described by one or more equations (a circle, a rectangle's sides, etc.). Extrema can occur here even when no interior critical point exists.
Step 1: Find interior critical points by setting fₓ = 0 and f_y = 0, solving for (x, y)
Step 2: Check the boundary
Parameterise each boundary segment (make the boundary a function of one variable)
Plug the boundary equation into f to reduce to a single-variable function
Take the derivative of that single-variable function, set it to zero, and solve
Pay attention to which values actually lie within the bounds of the boundary segment
Step 3: Evaluate f at every candidate point (interior critical points, boundary critical points, and corners or endpoints of boundary segments)
Step 4: The largest value is the absolute maximum; the smallest is the absolute minimum
Step 1: Write the squared distance from a general point on the surface to the given point (x₀, y₀, z₀)
distance² = (x - x₀)² + (y - y₀)² + (z - z₀)²
Step 2: Substitute z = f(x, y) (the surface equation) to eliminate z, leaving a function of x and y only
Step 3: Find fₓ and f_y of the squared-distance function and set both to zero
Step 4: Find the critical points and use the second derivative test to confirm you have a minimum (not a max or saddle)
Step 5: Plug the critical-point values of x and y back into z = f(x, y) to get z, then compute the actual distance
You minimise distance² rather than distance because it avoids square roots and gives the same critical points.
\text{distance}^2 = (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2Substitute the surface equation for z before differentiating.
Minimum distance problems appear in computer graphics (finding the closest point on a mesh to a cursor), in GPS positioning (closest point on a reference surface), and in manufacturing (tolerancing, where you need the shortest distance from a part surface to a datum plane).
Students often forget the boundary entirely and only report interior critical points. On a bounded domain, the absolute extrema might be on the boundary, not at an interior point.
A common error in distance problems is minimising the actual distance (with the square root) instead of distance². Both give the same critical points, but the algebra with the square root is far messier and leads to mistakes.
Students sometimes neglect to check that boundary critical points actually fall within the boundary segment. If a boundary runs from t = 0 to t = 2, a critical point at t = 3 is outside the domain and does not count.
Forgetting endpoints or corners of the boundary. These are always candidates for absolute extrema.
⚠️ Absolute extrema problems are a common exam question type. The full procedure (interior + boundary + endpoints, then compare) is expected.
⚠️ Show all candidate points and their f-values in a list or table. Examiners want to see you compared them systematically.
⚠️ Minimum distance problems may ask for the distance or the closest point. Read the question carefully to know which answer to give.
True or false: On a bounded domain, the absolute maximum must occur at an interior critical point. (False. It can occur on the boundary or at a corner.)
Fill in the blank: When checking a boundary, you reduce f to a function of ____ variable(s). (one)
True or false: In a minimum distance problem, you should minimise distance rather than distance². (False. Minimise distance² to avoid messy square roots; the critical points are the same.)
Q: Outline the full procedure for finding the absolute maximum of f(x, y) on a closed, bounded region.
A: Find all interior critical points (set fₓ = 0, f_y = 0). Parameterise each boundary segment and find critical points of f along each one. Evaluate f at all interior critical points, boundary critical points, and corners or endpoints. The largest value is the absolute maximum.
Q: Why do we minimise distance² instead of distance in minimum distance problems?
A: Because the square root in the distance formula complicates the derivatives without changing which point minimises it. Distance² and distance share the same critical points.
Q: A boundary is given by y = 4 - x² for 0 ≤ x ≤ 2. You find a critical point at x = 3. Is it valid?
A: No. x = 3 is outside the interval [0, 2], so it is not within the boundary and cannot be a candidate.
Absolute extrema builds on critical-point classification from the tangent planes and second derivative test topic. It also connects to Lagrange multipliers (if covered), which handle constrained optimisation without parameterising the boundary. The distance-minimisation technique reappears in least-squares fitting and in finding closest points on parametric surfaces.
Absolute extrema, global max, global min, bounded domain, closed region, boundary analysis, minimum distance to surface, closest point on surface, extreme value theorem, constrained optimisation, MATH 126, Calculus III, UW