Maxima and Minima of Multivariable Functions — Mathematics Knowledge System

Overview

For f(x,y), critical points occur where the gradient is zero (both f_x = 0 and f_y = 0) or is undefined. Critical points can be local maxima, local minima, or saddle points. The second derivative test (using the discriminant D = f_{xx}f_{yy} - f_{xy}^2) classifies them. Global extrema on closed bounded regions also occur on the boundary.

Intuition

Think of f(x,y) as a landscape (mountain/valley). Critical points are where the terrain is perfectly flat (gradient = 0). At a local max, you're at a peak. At a local min, you're in a valley. A saddle point looks like a mountain pass: it's a minimum in one direction and a maximum in another (like the middle of a saddle). The discriminant D tells you which case you're in.

Formal Definition

Definition

Critical points: gradient(f) = (0, 0). Second derivative test: D = f_{xx}f_{yy} - (f_{xy})^2 at (a,b). If D > 0 and f_{xx} > 0: local min. If D > 0 and f_{xx} < 0: local max. If D < 0: saddle point. If D = 0: inconclusive. The Hessian matrix H = [[f_{xx}, f_{xy}],[f_{yx}, f_{yy}]]; D = det(H).

\nabla f = \mathbf{0} \Rightarrow \text{critical point}

Critical point condition

D = f_{xx}f_{yy} - (f_{xy})^2 = \det(H)

Discriminant (Hessian determinant)

D > 0,; f_{xx} > 0 \Rightarrow \text{local min};\quad D > 0,; f_{xx} < 0 \Rightarrow \text{local max}

Classification

Notation

Notation	Meaning
$H_f$	Hessian matrix of f: second-order partial derivatives
$D = \det(H)$	Hessian determinant used in second derivative test

Theorems

Theorem 1: Extreme Value Theorem (multivariate)

A continuous function on a closed bounded set in R^n attains its absolute maximum and minimum. To find them: find all interior critical points and evaluate f there; evaluate f on the boundary; compare all values.

Worked Examples

1
f_x = 3x^2 - 3y = 0 and f_y = 3y^2 - 3x = 0.
$x^2 = y,\quad y^2 = x$
2
Substitute: x^4 = x => x(x^3 - 1) = 0, so x=0 or x=1. Points: (0,0) and (1,1).
$(0,0),;(1,1)$
3
Hessian: f_{xx}=6x, f_{yy}=6y, f_{xy}=-3. D=36xy-9.
4
At (0,0): D=0*0-9=-9<0. Saddle point.
$(0,0): \text{saddle}$
5
At (1,1): D=36-9=27>0, f_{xx}=6>0. Local minimum.
$(1,1): \text{local min, } f(1,1) = -1$

✓ Answer

(0,0) is a saddle; (1,1) is a local minimum with f(1,1) = -1.

Practice Problems

Mediumapplication

Find the minimum distance from the origin to the plane 2x + y + 2z = 9.

Common Mistakes

Common Mistake

Only checking interior critical points and ignoring the boundary.

On a closed bounded domain, absolute extrema can occur on the boundary. Check all interior critical points AND the boundary (often using single-variable techniques along the boundary curves).

Quiz

At a critical point with D < 0 (Hessian discriminant negative), the critical point is:

Local maximumLocal minimumSaddle pointInflection point

Historical Background

Optimization of functions of several variables developed alongside multivariable calculus in the 18th and 19th centuries. Lagrange (1788) introduced multipliers for constrained optimization. The second derivative test for multivariable functions uses the Hessian matrix, systematized in the 19th century. Today, optimization in high dimensions is central to machine learning and operations research.

1788
Lagrange introduces Lagrange multipliers for constrained optimization
Joseph-Louis Lagrange
1800s
The Hessian matrix and second-order conditions systematized

Summary

Critical points: gradient(f) = 0. Classify using D = f_{xx}f_{yy} - f_{xy}^2.
D > 0, f_{xx} > 0: local min. D > 0, f_{xx} < 0: local max. D < 0: saddle. D = 0: inconclusive.
Absolute extrema on closed bounded domain: compare interior critical points and boundary values.

References

BookStewart, J. Multivariable Calculus. 8th ed. Brooks/Cole, 2015.