Absolute Extrema on Closed Regions

Existence comes before calculation

A continuous real function on a nonempty compact region attains an absolute maximum and minimum. In Euclidean space, a closed bounded region is compact. If the domain is open, unbounded or contains a discontinuity, attainment needs a separate argument.

For a smooth function on a compact region with piecewise smooth boundary, check interior stationary points, optimize on each boundary piece, and include corners or endpoints. Boundary constraints reduce the available directions: the full gradient need not vanish at a boundary extremum.

Visual guide

VISUAL GUIDECheck the interior and the whole boundary

Boundary of the disk
A level circle of f

For f = x² + y² − 2x on the radius-2 disk, level circles are centered at (1, 0). The interior minimum is there. The farthest point of the disk from that center, (−2, 0), gives the maximum.

Worked example: a disk

For $f(x,y)=x^2+y^2-2x$ on $x^2+y^2\le4$ , the only interior stationary point is $(1,0)$ , where $f=-1$ . On the boundary, $x^2+y^2=4$ , so $f=4-2x$ . Since $-2\le x\le2$ , boundary values range from $0$ at $(2,0)$ to $8$ at $(-2,0)$ .

Comparing all candidates gives absolute minimum $-1$ at $(1,0)$ and maximum $8$ at $(-2,0)$ . Stopping after the interior point would miss the maximum entirely.

Worked example: a triangular boundary

For $f=xy$ on $x\ge0,y\ge0,x+y\le2$ , the coordinate edges give value zero. On the sloping edge $y=2-x$ , the function becomes $2x-x^2$ , whose maximum is $1$ at $x=1$ . The vertices also give zero.

There is no interior stationary point with positive coordinates because $\nabla f=(y,x)$ cannot vanish there. Hence the absolute maximum is $1$ at $(1,1)$ and the minimum is zero along both coordinate edges. Extrema need not occur at isolated points.

If a boundary is curved, parametrize it or use a regular constraint method. If a denominator or square root restricts the domain, identify excluded points before applying the compact-region theorem. A finite supremum at a missing boundary point is not necessarily an attained maximum.

Practice

Find extrema of $x+y$ on $[0,1]^2$ .
Does $f(x,y)=x$ attain a maximum on the open unit disk?
Where is the minimum of $x^2+y^2$ on $1\le x^2+y^2\le4$ ?

Show worked solutions

Minimum $0$ at $(0,0)$ ; maximum $2$ at $(1,1)$ .
No. Its supremum is $1$ , but the required point $(1,0)$ is excluded.
Value $1$ occurs everywhere on the inner circle, which is part of the boundary.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.