Absolute Extrema on Closed Regions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.7: “Maximum and Minimum Values” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-7-maximum-and-minimum-values |
| Textbook used in class | Stewart, Calculus, Section 14.7: “Maximum and Minimum Values” (Examples 5, 6) |
Quick Reference
Extreme Value Theorem (2D): If $f$ is continuous on a closed, bounded region $D$, then $f$ attains an absolute maximum and an absolute minimum on $D$.
Algorithm for absolute extrema:
- Find all critical points of $f$ inside $D$ and evaluate $f$ there.
- Find the extreme values of $f$ on each piece of the boundary of $D$.
- The absolute maximum is the largest of all values found; the absolute minimum is the smallest.
Motivation
The single-variable Extreme Value Theorem says: a continuous function on a closed interval $[a,b]$ achieves its absolute maximum and minimum somewhere on $[a,b]$. The two-variable version says the same thing for a closed bounded region (like a disk or a rectangle): the absolute extrema are achieved somewhere, either in the interior or on the boundary.
The search is systematic: check the interior (via critical points and the second derivative test) and then parametrize the boundary to find extrema there.
Key Concepts
1. What is a Closed Bounded Region?
A region is closed if it contains its boundary (for example, a disk $x^2 + y^2 \leq 4$ is closed; the open disk $x^2 + y^2 < 4$ is not). A region is bounded if it fits inside some large circle.
For closed bounded regions, the Extreme Value Theorem guarantees that absolute extrema exist.
2. Boundary Parametrization
The boundary of a 2D region is typically a curve. To find the extrema of $f$ on the boundary, parametrize the boundary as $\mathbf{r}(t)$, substitute into $f$, and use single-variable calculus to optimize the resulting function of $t$.
For a rectangle, the boundary consists of four line segments, each handled separately.
Worked Example
Find the absolute maximum and minimum of $f(x,y) = x^2 - 2xy + 2y$ on the triangle with vertices $(0,0)$, $(3,0)$, $(3,3)$. (Adapted from Stewart 14.7.)
Step 1: Interior critical points.
$f_x = 2x - 2y = 0 \implies x = y$.
$f_y = -2x + 2 = 0 \implies x = 1$.
So $x = y = 1$. Check: $(1,1)$ is inside the triangle. $f(1,1) = 1 - 2 + 2 = 1$.
Step 2: Boundary.
The triangle has three edges.
Edge 1: $y = 0$, $0 \leq x \leq 3$. $f(x,0) = x^2$. Maximum $9$ at $x=3$; minimum $0$ at $x=0$.
Edge 2: $x = 3$, $0 \leq y \leq 3$. $f(3,y) = 9 - 6y + 2y = 9 - 4y$. Decreasing in $y$; maximum $9$ at $y=0$, minimum $9-12=-3$ at $y=3$.
Edge 3: $y = x$, $0 \leq x \leq 3$. $f(x,x) = x^2 - 2x^2 + 2x = -x^2 + 2x$. Vertex at $x=1$: $f = 1$. Endpoints: $f(0,0) = 0$, $f(3,3) = -9+6 = -3$.
Step 3: Compare all values.
| Point | Value |
|---|---|
| $(1,1)$ interior | $1$ |
| $(0,0)$ | $0$ |
| $(3,0)$ | $9$ |
| $(3,3)$ | $-3$ |
Absolute maximum: $9$ at $(3,0)$. Absolute minimum: $-3$ at $(3,3)$.
the absolute maximum must occur at a critical point. The absolute maximum occurs at the point (interior or boundary) where $f$ is largest. Boundary points are not critical points of $f$ (as a function on the plane), yet the absolute extremum can occur there -- and often does. Checking only interior critical points is the most common error in this type of problem.
Leveled Practice
Problem 1. Find the absolute max and min of $f(x,y) = x^2 + y^2 - 2x$ on the closed disk $x^2+y^2 \leq 4$.
Show answer
Interior: $f_x = 2x-2 = 0$, $f_y = 2y = 0$ gives the critical point $(1,0)$. Check: $1+0 < 4$, so it is inside the disk. $f(1,0) = 1+0-2 = -1$.
Boundary: $x = 2\cos t$, $y = 2\sin t$.
$f = 4\cos^2 t + 4\sin^2 t - 4\cos t = 4 - 4\cos t$.
Max when $\cos t = -1$: $t = \pi$, point $(-2, 0)$, $f = 8$.
Min when $\cos t = 1$: $t = 0$, point $(2, 0)$, $f = 0$.
Compare: $-1$ (interior), $8$ (boundary), $0$ (boundary).
Absolute maximum: $8$ at $(-2,0)$. Absolute minimum: $-1$ at $(1,0)$.