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Extrema and Constrained Extrema

Extrema of Multivariable Functions

Let $z = f(x, y)$ be defined on a neighborhood of $P_0(x_0, y_0)$ . If for every point $P(x, y) \neq P_0$ in that neighborhood,

$f(x, y) < f(x_0, y_0) \quad \text{(or } f(x, y) > f(x_0, y_0) \text{)}$

then $f$ has a local maximum (or local minimum) at $P_0$ , and $P_0$ is called a maximum point (or minimum point). Maxima and minima together are called extrema.

Note: an extremum is a local concept—it compares with values near the point only, not global maxima/minima over the whole domain. A function may have several maxima and minima, and a local maximum need not exceed a local minimum.

Necessary Condition — Fermat’s Theorem

Theorem1

Fermat’s theorem (necessary condition): if $z = f(x, y)$ has partial derivatives at $(x_0, y_0)$ and attains an extremum there, then

$f_x(x_0, y_0) = 0, \quad f_y(x_0, y_0) = 0$

For an $n$ -variable function, all first partials vanish at an extremum:

$f_{x_i}(P_0) = 0 \quad (i = 1, 2, \ldots, n)$

Cautions:

The necessary condition requires the partial derivatives to exist. If they do not, the function can still attain an extremum—e.g. $f(x,y) = \sqrt{x^2+y^2}$ has a minimum at $(0,0)$ where the partials do not exist.
Vanishing partials are necessary but not sufficient: a point with zero gradient need not be an extremum (it may be a saddle point).

Critical Points and Stationary Points

A point $(x, y)$ where both first partial derivatives vanish, i.e.

$f_x(x, y) = 0, \quad f_y(x, y) = 0$

is called a stationary point (or critical point). More generally, any point with $\nabla f = \vec{0}$ is a stationary point.

Relation between extrema and stationary points:

A differentiable extremum point is always a stationary point (necessary condition).
A stationary point need not be an extremum (it may be a saddle point).
An extremum point may fail to be stationary if partials do not exist.

So candidate points are: (1) stationary points ( $\nabla f = \vec{0}$ ), and (2) points where partial derivatives do not exist. Then use a sufficient condition (second-derivative test) or the definition to decide.

Saddle Point

A stationary point that is neither a local maximum nor a local minimum is called a saddle point—the surface goes up along some directions and down along others, like a horse’s saddle.

Second-Derivative Test

The Hessian matrix of $z = f(x, y)$ at a stationary point is

$H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}$

Theorem2

Second-derivative test: let $(x_0, y_0)$ be a stationary point, and set

$A = f_{xx}(x_0,y_0), \quad B = f_{xy}(x_0,y_0), \quad C = f_{yy}(x_0,y_0), \quad \Delta = AC - B^2$

If $\Delta > 0$ and $A > 0$ : local minimum.
If $\Delta > 0$ and $A < 0$ : local maximum.
If $\Delta < 0$ : saddle point.
If $\Delta = 0$ : the test is inconclusive.

Worked Examples: Unconstrained Extrema

Example 1: Second-Derivative Test

Find the extrema of $f(x, y) = x^2 + y^2$ .

Reference Answer

extremastationary pointsecond test

Approach:

Classify a stationary point by evaluating the second partial derivatives and the discriminant built from them.

Steps:

Stationary point: $f_x = 2x = 0$ , $f_y = 2y = 0$ , so $(0,0)$ .
Second derivatives: $f_{xx}=2, f_{xy}=0, f_{yy}=2$
$\Delta = 4 - 0 = 4 > 0$ , and $A = 2 > 0$ : local minimum.

Answer: local (and global) minimum $0$ at $(0,0)$ .

Example 2: Saddle Point

Find the extrema of $f(x, y) = x^2 - y^2$ .

Reference Answer

extremasaddle pointsecond test

Approach:

A negative discriminant means the Hessian is indefinite, so the point is a saddle rather than an extremum.

Steps:

Stationary point: $f_x = 2x = 0$ , $f_y = -2y = 0$ , so $(0,0)$ .
$f_{xx}=2, f_{xy}=0, f_{yy}=-2$
$\Delta = -4 < 0$ : saddle point.

Answer: $(0,0)$ is a saddle point—not an extremum.

Constrained Extrema and Lagrange Multipliers

Finding the extremum of $z = f(x, y)$ (the objective function) subject to the constraint $\varphi(x, y) = 0$ is called a constrained extremum problem. In general:

$\text{extreme } f(x_1, \ldots, x_n) \quad \text{subject to } \varphi_i(x_1, \ldots, x_n) = 0 \quad (i = 1, \ldots, m)$

Unconstrained vs constrained:

Unconstrained: the variables range freely over the domain.
Constrained: the variables must satisfy constraint equations; the extremum is confined to the constraint surface.

Two solution methods:

Substitution: solve the constraint for some variables and substitute into the objective function to reduce to an unconstrained problem. Fails when the constraint is complicated.
Lagrange multipliers: introduce multipliers to turn the problem into a system of equations. This is the general method.

Lagrange Multiplier Method (Single Constraint)

Theorem3

To find possible extrema of $f(x, y)$ subject to $\varphi(x, y) = 0$ , form the Lagrangian:

$L(x, y, \lambda) = f(x, y) + \lambda \varphi(x, y)$

where $\lambda$ is the Lagrange multiplier. Set all first partials to zero:

$\begin{cases} L_x = f_x + \lambda \varphi_x = 0 \\ L_y = f_y + \lambda \varphi_y = 0 \\ L_\lambda = \varphi(x, y) = 0 \end{cases}$

Solving this system gives the candidate constrained extremum points.

Intuition: at a constrained extremum, the gradient of the objective function is parallel to the gradient of the constraint— $\nabla f = -\lambda \nabla \varphi$ —so the level curves of $f$ are tangent to the constraint curve.

Example 3: Lagrange Multipliers

Maximize $f(x, y) = x + y$ subject to $x^2 + y^2 = 2$ .

Reference Answer

constrained extremumLagrange multipliercircle

Approach:

To extremize a function subject to a constraint, build the Lagrangian and set all of its partial derivatives to zero together with the constraint equation.

Steps:

Lagrangian: $L = x + y + \lambda(x^2 + y^2 - 2)$
Set partials to zero: $L_x = 1 + 2\lambda x = 0$ $L_y = 1 + 2\lambda y = 0$ $L_\lambda = x^2 + y^2 - 2 = 0$
From the first two: $x = y = -\frac{1}{2\lambda}$
Substitute: $2\left(-\frac{1}{2\lambda}\right)^2 = 2 \Rightarrow \lambda^2 = \frac14 \Rightarrow \lambda = \pm\frac12$
$\lambda = -\frac12$ gives $x=y=1$ ; $\lambda = \frac12$ gives $x=y=-1$
$f(1,1) = 2$ , $f(-1,-1) = -2$

Answer: maximum $2$ at $(1,1)$ ; minimum $-2$ at $(-1,-1)$ .

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$f_x, f_y$	Partial derivative	f sub x, f sub y	First partial derivatives
$A, B, C$	Number	A, B, C	$f_{xx}, f_{xy}, f_{yy}$ at a stationary point
$\Delta$	Greek letter	Delta	The discriminant $AC - B^2$
$\lambda$	Greek letter	Lambda	The Lagrange multiplier
$L$	Function	L	The Lagrangian function
$\varphi$	Greek letter	phi	The constraint function
$\nabla f$	Vector	nabla f	The gradient of $f$

Applications of the Gradient

The Extrema of Multivariable Functions

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Extrema and Constrained Extrema

Extrema of Multivariable Functions

Necessary Condition — Fermat’s Theorem

Critical Points and Stationary Points

Saddle Point

Second-Derivative Test

Worked Examples: Unconstrained Extrema

Example 1: Second-Derivative Test

Example 2: Saddle Point

Constrained Extrema and Lagrange Multipliers

Lagrange Multiplier Method (Single Constraint)

Example 3: Lagrange Multipliers

Summary

Symbols Used in This Article

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