Theory

Unconstrained and constrained maxima and minima

How to find and classify the maxima and minima of a function of two variables, with stationary points and the Hessian matrix, how to look for absolute extrema on a closed and bounded set and extrema on a constraint, with substitution and with Lagrange multipliers, and how these methods solve the problems of profit, of the consumer and of minimum cost.

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Relative and absolute maxima and minima
Stationary points and the first-order condition

Relative and absolute maxima and minima

Let $f$ be a function of two variables defined on a set $D \subseteq R^{2}$ and let $P_{0} = (x_{0}, y_{0}) \in D$ . The point $P_{0}$ is a relative maximum point (or local maximum point) of $f$ if there exists a disk $B$ centered at $P_{0}$ such that

f (x, y) \leq f (x_{0}, y_{0}) for every (x, y) \in B \cap D,

and it is a relative minimum point if the opposite inequality holds, $f (x, y) \geq f (x_{0}, y_{0})$ . If the inequality holds for every $(x, y) \in D$ , the maximum or the minimum is called absolute. With strict inequalities for $(x, y) \neq = (x_{0}, y_{0})$ one speaks of a strict maximum or minimum. Maxima and minima are together called extrema.

In other words, $P_{0}$ is a relative minimum point when no nearby point gives $f$ a smaller value, and an absolute minimum point when no point of the domain does. For example $f (x, y) = x^{2} + y^{2}$ has a strict absolute minimum at $(0, 0)$ , because $f (0, 0) = 0$ and $x^{2} + y^{2} > 0$ for every $(x, y) \neq = (0, 0)$ . The notions of disk, interior point and closed and bounded set are those of the topic Functions of two variables.

Stationary points and the first-order condition

1. Fermat's theorem in two variables

Let $P_{0} = (x_{0}, y_{0})$ be an interior point of the domain of $f$ that is a relative maximum or minimum point. If the partial derivatives exist at $P_{0}$ , then

{f_{x} (x_{0}, y_{0}) = 0 f_{y} (x_{0}, y_{0}) = 0 that is \nabla f (x_{0}, y_{0}) = (0, 0) .

The theorem reduces the search for interior extrema to solving a system of two equations in two unknowns. As in the one-variable case, the condition is necessary but not sufficient, and the points found must then be classified.

Proof

The idea is to look at $f$ only along the horizontal line through $P_{0}$ , where it becomes a function of one variable. Set $g (t) = f (x_{0} + t, y_{0})$ . Since $P_{0}$ is interior, $g$ is defined on an interval that contains $t = 0$ in its interior. If $P_{0}$ is a relative maximum point of $f$ , then $g (t) \leq g (0)$ for $t$ near $0$ , so $t = 0$ is a relative maximum point of $g$ . The derivative $g^{'} (0)$ is the limit of the difference quotient $\frac{f ( x _{0} + t , y _{0} ) - f ( x _{0} , y _{0} )}{t}$ , that is, by definition $f_{x} (x_{0}, y_{0})$ . Fermat’s theorem applied to $g$ therefore gives $f_{x} (x_{0}, y_{0}) = 0$ . With the vertical line, $h (t) = f (x_{0}, y_{0} + t)$ , one obtains in the same way $f_{y} (x_{0}, y_{0}) = 0$ . The case of the minimum is analogous.

2. Stationary point

An interior point $P_{0}$ at which $\nabla f (P_{0}) = (0, 0)$ is called a stationary point (or critical point) of $f$ . For a function with partial derivatives at every point of an open domain, the relative maxima and minima must be sought only among the stationary points. For example, for $f (x, y) = x^{2} + y^{2} - 2 x + 4 y$ we have

{f_{x} = 2 x - 2 = 0 f_{y} = 2 y + 4 = 0 ⟺ {x = 1 y = - 2

so the only stationary point is $(1, - 2)$ . Completing the squares one writes $f (x, y) = (x - 1)^{2} + (y + 2)^{2} - 5$ . The sum of the two squares is positive at every point other than $(1, - 2)$ , so $(1, - 2)$ is an absolute minimum with value $- 5$ .

3. Saddle point

A stationary point $P_{0}$ is a saddle point if every disk centered at $P_{0}$ contains points where $f > f (P_{0})$ and points where $f < f (P_{0})$ . A saddle point is therefore neither a relative maximum point nor a relative minimum point. The typical example is $f (x, y) = x^{2} - y^{2}$ at the origin. The gradient $(2 x, - 2 y)$ vanishes at $(0, 0)$ , but along the $x$ -axis the function equals $x^{2}$ , positive for $x \neq = 0$ , and along the $y$ -axis it equals $- y^{2}$ , negative for $y \neq = 0$ . The name comes from the shape of the graph near the origin, similar to a saddle, with heights that increase when moving along the $x$ -axis and decrease when moving along the $y$ -axis.

z = f(t, 0) = t^2

z = f(0, t) = -t^2

Restrictions of

f (x, y) = x^{2} - y^{2}

to the two axes: at

t = 0

the first has a minimum and the second a maximum, so

(0, 0)

is a saddle point

The theory continues with 6 more sections (the hessian matrix and the hessian test, inconclusive cases, absolute maxima and minima on closed and bounded sets…), plus all worked examples, the solutions and the PDF.

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