[CALCULUS] Global extrema

The extreme-value theorem and a complete interior-and-boundary workflow for absolute extrema.

This page finds the absolute maximum and minimum of a function on a region. Unlike a local test, an absolute extremum compares every point of the domain, so the boundary of the region enters the calculation. On a closed and bounded region the extreme-value theorem guarantees that both extrema exist; interior candidates come from the stationary points and each boundary piece gives a lower-dimensional optimisation problem.

Closed and bounded regions

[DEFINITION]

Closed and bounded region

A region $D\subseteq\mathbb{R}^2$ is closed if it contains its boundary $\partial D$ and bounded if it fits inside a sufficiently large disk: there is an $R>0$ such that $\|\mathbf{x}\|\le R$ for every $\mathbf{x}\in D$ .

An interior point has a neighbourhood contained entirely in $D$ . A boundary point has every neighbourhood meeting both $D$ and its complement; the set of boundary points is written $\partial D$ .

The closed unit disk

\{(x,y):x^2+y^2\le1\}

contains its boundary, the unit circle $x^2+y^2=1$ . An open disk omits that circle and is not closed.

[1]:The multivariable extreme-value theorem

[THEOREM]

The multivariable extreme-value theorem

If $f$ is continuous on a nonempty closed and bounded region $D\subseteq\mathbb{R}^2$ , then $f$ attains an absolute maximum and an absolute minimum on $D$ .

An absolute extremum can occur at an interior critical point or anywhere on the boundary $\partial D$ . The theorem guarantees existence, but it does not identify the candidates. The boundary is usually an infinite set, so it must be treated as a lower-dimensional optimisation problem.

If either closedness or boundedness fails, an extremum need not be attained; for example, $f(x,y)=x$ on the open unit disk has supremum $1$ but no maximum.

The candidate workflow

[METHOD]

Find absolute extrema on a closed bounded region

For a continuous $f$ on a nonempty closed and bounded region $D$ , use the following workflow, checking differentiability wherever a derivative is used:

find all stationary points in the differentiable interior of $D$ and separately record any interior points where $f$ is not differentiable;
restrict $f$ to each boundary piece;
find critical points of every restricted function, including endpoints and corners;
evaluate $f$ at every candidate;
compare the resulting values.

Neither part of the search can be skipped: a boundary point can have a larger or smaller value than every interior critical point and a corner may not be detected by a one-variable derivative.

[METHOD]

Analyse a boundary by parameterisation

To analyse a differentiable boundary curve, parameterise it as

\mathbf{r}(t)=(x(t),y(t))

and optimise the one-variable function

\phi(t)=f(x(t),y(t)).

Include all parameter endpoints when the boundary is a segment or a piecewise-smooth curve and include corners separately.

Parameterisation converts the boundary part of a two-variable problem into one-variable calculus. When the boundary is naturally a level curve $g=0$ , Lagrange multipliers solve the boundary problem without a parameterisation.

[EXAMPLE]

Extrema on a rectangle

Find the maximum and minimum of

f(x,y)=x^2-x^2y+3y^2

D=\{(x,y):-1\le x\le1,\ -1\le y\le1\}.

For interior candidates,

f_x=2x(1-y), \qquad f_y=-x^2+6y.

The equations $f_x=f_y=0$ give either $x=0$ , hence $y=0$ , or $y=1$ , hence $x^2=6$ , which is outside the rectangle. The only interior candidate is $(0,0)$ , with

f(0,0)=0.

On the vertical edges $x=\pm1$ ,

f(\pm1,y)=1-y+3y^2.

The derivative is $-1+6y$ , so the edge critical point is $y=1/6$ , where

f\left(\pm1,\frac16\right) =1-\frac16+3\left(\frac16\right)^2 =\frac{11}{12}.

The edge endpoints give

f(\pm1,-1)=5, \qquad f(\pm1,1)=3.

On the horizontal edges,

f(x,1)=3, \qquad f(x,-1)=2x^2+3.

The latter has minimum $3$ at $x=0$ and maximum $5$ at $x=\pm1$ . Comparing all candidates gives

{\min_D f=0\text{ at }(0,0)}

and

{\max_D f=5\text{ at }(1,-1)\text{ and }(-1,-1)}.

The minimum is interior, while the maximum occurs at corners.

[exam]

[1]:Find absolute extrema on a closed bounded region

[EXAMPLE]

Extrema on the closed unit disk

Let

f(x,y)=x^2+4xy+y^2

D=\{(x,y):x^2+y^2\le1\}.

The interior gradient is

\nabla f = \begin{pmatrix} 2x+4y\\ 4x+2y \end{pmatrix}.

Solving $\nabla f=\mathbf{0}$ gives $(0,0)$ , with $f(0,0)=0$ .

The boundary is the unit circle. Parameterise it by

x=\cos t, \qquad y=\sin t.

Then

\begin{aligned} f(\cos t,\sin t) &=\cos^2t+4\cos t\sin t+\sin^2t\\ &=1+2\sin(2t). \end{aligned}

Since $-1\le\sin(2t)\le1$ ,

-1\le f(\cos t,\sin t)\le3.

The maximum $3$ occurs at

\left(\frac1{\sqrt2},\frac1{\sqrt2}\right), \qquad \left(-\frac1{\sqrt2},-\frac1{\sqrt2}\right),

and the minimum $-1$ occurs at

\left(\frac1{\sqrt2},-\frac1{\sqrt2}\right), \qquad \left(-\frac1{\sqrt2},\frac1{\sqrt2}\right).

Comparing with the interior value gives

{\max_D f=3}

at the two same-sign boundary points and

{\min_D f=-1}

at the two opposite-sign boundary points.

[exam]

The boundary calculation can also be done without a parameter by treating $x^2+y^2=1$ as a constraint. The constrained extrema page develops this method, in which a gradient-parallel condition replaces the parameter derivative.

Global extrema

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