Absolute Extrema on Regions, multivariable and vector calculus lesson

A metal plate is heated unevenly and you want its hottest and coldest spots. The answer could sit in the middle of the plate, somewhere along an edge, or right at a corner, and a search that skips any of these can miss it. This lesson gives a guarantee that the extremes exist and a search that looks in every place they can hide.

Key terms

Continuous function on a closed bounded region

Definition Let $D$ be a set in $\mathbb{R}^2$ . A point $P$ is a boundary point of $D$ if every disk centered at $P$ contains points of $D$ and points not in $D$ . The set $D$ is closed if it contains all of its boundary points, and bounded if it lies inside some disk. A number $f(P)$ is the absolute maximum of $f$ on $D$ if $f(P) \ge f(Q)$ for every $Q$ in $D$ , and the absolute minimum if $f(P) \le f(Q)$ for every $Q$ in $D$ .

The extreme value theorem says the following. If $D$ is a nonempty, closed and bounded set in $\mathbb{R}^2$ and $f$ is continuous at every point of $D$ , then there are points $P$ and $Q$ in $D$ with

\begin{gathered} f(Q) \le f(X) \le f(P) \\[0.4em] \text{for all } X \text{ in } D. \end{gathered}

The same holds in $\mathbb{R}^n$ . Closed and bounded sets are also called compact.

In plain words If the region includes its edge, does not run off to infinity, and the function has no jumps, then a highest and a lowest value really exist. The theorem promises they exist but does not say where.

Example Take the closed disk $D$ given by $x^2 + y^2 \le 4$ and the continuous function $f(x, y) = x^2 + y^2$ . The minimum is $0$ , reached only at the origin. The maximum is $4$ , and it is reached at every point of the circle $x^2 + y^2 = 4$ . So an absolute maximum can occur at infinitely many points at once.

Interior critical points

Definition A point $P$ of $D$ is an interior point if some disk centered at $P$ lies entirely in $D$ . If $f$ has a local maximum or local minimum at an interior point $P$ and both first partial derivatives exist there, then

f_x(P) = 0 \quad \text{and} \quad f_y(P) = 0.

An interior critical point is an interior point where both partial derivatives are zero or where at least one of them does not exist. An absolute extremum that occurs in the interior of $D$ is also a local extremum, so it must occur at an interior critical point.

In plain words Inside the region, the only places a high or low point can sit are where the surface is flat or where it has a crease or spike. You do not need to classify these points. You only record the value of $f$ at each one.

Example Throughout this lesson, let $f(x, y) = x^2 - 2xy + 2y$ on the rectangle $R = [0, 3] \times [0, 2]$ . Its partial derivatives are

f_x = 2x - 2y, \qquad f_y = -2x + 2.

Setting both equal to zero gives $x = 1$ and then $y = 1$ . The point $(1, 1)$ lies inside $R$ , and $f(1, 1) = 1 - 2 + 2 = 1$ . A critical point outside the region would simply be ignored.

Boundary curves

Definition Suppose the boundary of $D$ is made of finitely many smooth curves $C_1, \dots, C_k$ , each given by a parametrization $\mathbf{r}_i(t)$ with $t$ in a closed interval $[\alpha_i, \beta_i]$ . On $C_i$ the values of $f$ are those of the one-variable function

g_i(t) = f(\mathbf{r}_i(t)), \qquad \alpha_i \le t \le \beta_i.

The extreme values of $f$ on $C_i$ are the extreme values of $g_i$ on $[\alpha_i, \beta_i]$ , found as in single-variable calculus from the critical numbers of $g_i$ in $(\alpha_i, \beta_i)$ and the two endpoints. A point where $f$ has an extremum on the boundary need not have $\nabla f = 0$ .

In plain words Walk along each edge, one at a time. On an edge only one quantity varies, so the problem turns into an ordinary closed interval problem.

Example The rectangle $R$ has four edges.

Edge	Values of $f$	Smallest	Largest
$y = 0$ , $0 \le x \le 3$	$x^2$	$0$ at $x = 0$	$9$ at $x = 3$
$x = 3$ , $0 \le y \le 2$	$9 - 4y$	$1$ at $y = 2$	$9$ at $y = 0$
$y = 2$ , $0 \le x \le 3$	$(x - 2)^2$	$0$ at $x = 2$	$4$ at $x = 0$
$x = 0$ , $0 \le y \le 2$	$2y$	$0$ at $y = 0$	$4$ at $y = 2$

On the top edge the function $x^2 - 4x + 4$ has a critical number at $x = 2$ , which gives the point $(2, 2)$ where $f = 0$ . At $(2, 2)$ the gradient is $\langle 0, -2 \rangle$ , not zero, yet $f$ is smallest there along that edge.

Corners

Definition A corner of the boundary is a point where two of the smooth boundary curves meet, or more generally a point of the boundary where it fails to be smooth. In the parametrizations above, corners are endpoints of the intervals $[\alpha_i, \beta_i]$ . The value of $f$ at every corner belongs on the list of candidates.

The closed region method puts the pieces together. For $f$ continuous on a closed bounded region $D$ ,

evaluate $f$ at every interior critical point,
find the extreme values of $f$ on every boundary curve, including its endpoints,
evaluate $f$ at every corner.

The largest value found is the absolute maximum and the smallest is the absolute minimum.

In plain words Corners are where one edge hands over to the next, and endpoints of a closed interval are always candidates. Many absolute extrema on rectangles and triangles sit exactly there.

Example The corners of $R$ give $f(0, 0) = 0$ , $f(3, 0) = 9$ , $f(3, 2) = 1$ and $f(0, 2) = 4$ . Here is the full list.

Point	Type	Value
$(1, 1)$	interior critical point	$1$
$(2, 2)$	edge critical number	$0$
$(0, 0)$	corner	$0$
$(3, 0)$	corner	$9$
$(3, 2)$	corner	$1$
$(0, 2)$	corner	$4$

The absolute maximum is $9$ at the corner $(3, 0)$ . The absolute minimum is $0$ , reached at $(0, 0)$ and at $(2, 2)$ . The interior critical point is neither.

Constrained domains

Definition A constrained domain is a region $D$ described by inequality constraints, the set of all points $(x, y)$ with

\begin{gathered} G_1(x, y) \le c_1, \quad \dots, \\[0.4em] G_k(x, y) \le c_k, \end{gathered}

where every $G_i$ is continuous on $\mathbb{R}^2$ . Finding the extrema of $f$ on $D$ means optimizing $f$ subject to these constraints. Such a set is always closed, but whether it is bounded must be checked. A point where every inequality is strict is an interior point, so every boundary point makes at least one constraint an equation. The boundary curves lie on the level curves $G_i(x, y) = c_i$ , and corners are typically points where two constraints hold with equality at once. On a boundary curve the variables are tied together, so $f$ is optimized along it by parametrizing the curve, by solving the equation for one variable while keeping the allowed range of the other, or by Lagrange multipliers. The function $f$ must also be continuous on all of $D$ , so $D$ has to lie inside the domain of $f$ .

In plain words The edge of a disk or an ellipse is a single curved boundary. Substituting the constraint is fine, as long as you remember which values the remaining variable may take.

Example Find the extreme values of $f(x, y) = x^2 + 2y^2 - x$ on the closed disk $x^2 + y^2 \le 1$ , a constrained domain with the single constraint $G(x, y) = x^2 + y^2 \le 1$ . Inside, $f_x = 2x - 1$ and $f_y = 4y$ vanish only at $\left(\tfrac{1}{2}, 0\right)$ , an interior point with $f = -\tfrac{1}{4}$ . On the circle $y^2 = 1 - x^2$ , and here $x$ is restricted to $[-1, 1]$ , so

\begin{aligned} f &= x^2 + 2(1 - x^2) - x \\[0.4em] &= 2 - x - x^2, \qquad -1 \le x \le 1. \end{aligned}

Its derivative $-1 - 2x$ is zero at $x = -\tfrac{1}{2}$ , which gives the points $\left(-\tfrac{1}{2}, \pm\tfrac{\sqrt{3}}{2}\right)$ with $f = \tfrac{9}{4}$ . The endpoints $x = 1$ and $x = -1$ give $f(1, 0) = 0$ and $f(-1, 0) = 2$ . The absolute maximum is $\tfrac{9}{4}$ at $\left(-\tfrac{1}{2}, \pm\tfrac{\sqrt{3}}{2}\right)$ and the absolute minimum is $-\tfrac{1}{4}$ at $\left(\tfrac{1}{2}, 0\right)$ . Forgetting the range $[-1, 1]$ would let $2 - x - x^2$ run down to $-\infty$ , which is nonsense on a circle.

Open and unbounded regions

Definition If $D$ is not closed or not bounded, the extreme value theorem does not apply and a continuous function may have no absolute maximum, no absolute minimum, or neither. Existence then needs a separate argument. One standard tool is the following. Let $D$ be closed and nonempty, let $f$ be continuous on $D$ , and suppose $f(x, y) \to \infty$ as $\sqrt{x^2 + y^2} \to \infty$ with $(x, y)$ in $D$ . Then $f$ has an absolute minimum on $D$ . To find it, pick a point $P_0$ of $D$ and a radius $\rho$ so large that $P_0$ lies in the disk of radius $\rho$ about the origin and $f > f(P_0)$ at every point of $D$ outside that disk. The part of $D$ inside the closed disk is closed, bounded and nonempty, and its minimum is the absolute minimum on all of $D$ .

In plain words Without a closed edge, a value can be approached forever without being reached, and without a bounded region the values can escape to infinity. If the function grows large far away, the smallest value must be found somewhere in the middle.

Example On the open disk $x^2 + y^2 < 1$ the function $f(x, y) = x^2 + y^2$ has minimum $0$ at the origin but no maximum, since its values come as close to $1$ as you like and never equal $1$ .

For an unbounded case, take $f(x, y) = x^4 + y^4 - 4xy$ on all of $\mathbb{R}^2$ . With $r = \sqrt{x^2 + y^2}$ , the inequalities $x^4 + y^4 \ge \tfrac{1}{2} r^4$ and $4 \lvert xy \rvert \le 2r^2$ give

f(x, y) \ge \tfrac{1}{2} r^4 - 2r^2 = \tfrac{1}{2} r^2 (r^2 - 4),

so $f \ge 0$ when $r \ge 2$ . Since $f(1, 1) = -2$ , the minimum of $f$ on the closed disk $r \le 2$ is negative. It cannot occur on the circle $r = 2$ , where $f \ge 0$ , so it occurs at an interior point of the disk, which must be a critical point. Outside the disk $f \ge 0$ as well, so this is the minimum on the whole plane. Solving $4x^3 - 4y = 0$ and $4y^3 - 4x = 0$ gives $(0, 0)$ , $(1, 1)$ and $(-1, -1)$ , with values $0$ , $-2$ and $-2$ . The absolute minimum on $\mathbb{R}^2$ is $-2$ , at $(1, 1)$ and $(-1, -1)$ . There is no absolute maximum, because $f(x, 0) = x^4$ grows without bound.

Check yourself

Find the absolute extreme values of $f(x, y) = x^2 + y^2 - x - y$ on the closed triangle with corners $(0, 0)$ , $(2, 0)$ and $(0, 2)$ .

Show the answer

The only critical point is $\left(\tfrac{1}{2}, \tfrac{1}{2}\right)$ , which is inside, with $f = -\tfrac{1}{2}$ . On the edge $y = 0$ the values are $x^2 - x$ for $0 \le x \le 2$ , with smallest value $-\tfrac{1}{4}$ and largest $2$ , and the edge $x = 0$ is the same with the roles swapped. On the edge $y = 2 - x$ the values are $2(x - 1)^2$ for $0 \le x \le 2$ , between $0$ and $2$ . The corners give $0$ , $2$ and $2$ . The absolute maximum is $2$ at $(2, 0)$ and $(0, 2)$ , and the absolute minimum is $-\tfrac{1}{2}$ at $\left(\tfrac{1}{2}, \tfrac{1}{2}\right)$ .

Find the absolute extreme values of $f(x, y) = x + y$ on the annulus $1 \le x^2 + y^2 \le 4$ .

Show the answer

The gradient $\langle 1, 1 \rangle$ is never zero, so there are no interior critical points. The annulus has two boundary circles. On the outer one, $x = 2\cos t$ and $y = 2\sin t$ give $2(\cos t + \sin t)$ , which ranges from $-2\sqrt{2}$ to $2\sqrt{2}$ . On the inner circle the values range only from $-\sqrt{2}$ to $\sqrt{2}$ . The absolute maximum is $2\sqrt{2}$ at $(\sqrt{2}, \sqrt{2})$ and the absolute minimum is $-2\sqrt{2}$ at $(-\sqrt{2}, -\sqrt{2})$ .

Does the function $f(x, y) = e^{-(x^2 + y^2)}$ have an absolute maximum and an absolute minimum on $\mathbb{R}^2$ ?

Show the answer

Since $x^2 + y^2 \ge 0$ , every value satisfies $0 < f(x, y) \le 1$ , and $f(0, 0) = 1$ . So the absolute maximum is $1$ at the origin. There is no absolute minimum, because the values approach $0$ far from the origin but are never equal to $0$ . The plane is not bounded, so the extreme value theorem gives no guarantee.

Absolute Extrema on Regions

Key terms

Continuous function on a closed bounded region

Interior critical points

Boundary curves

Corners

Constrained domains

Open and unbounded regions

Check yourself

How this lesson connects