Critical Points of Multivariable Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.7: “Maximum and Minimum Values” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-7-maximum-and-minimum-values |
| Textbook used in class | Stewart, Calculus, Section 14.7: “Maximum and Minimum Values” (Examples 1, 2) |
Quick Reference
Critical point: $(a,b)$ is a critical point of $f$ if $f_x(a,b) = 0$ and $f_y(a,b) = 0$, or if one of the partial derivatives does not exist.
Necessary condition: If $f$ has a local max or min at $(a,b)$ and the partial derivatives exist there, then $(a,b)$ is a critical point.
Types of critical points: local maximum, local minimum, saddle point (neither max nor min).
Motivation
In single-variable calculus, the candidates for extreme values are points where $f'(x) = 0$ or $f'$ does not exist. In two variables, the gradient must vanish: $\nabla f = \mathbf{0}$, meaning BOTH $f_x = 0$ AND $f_y = 0$. A critical point might be a local maximum, a local minimum, or a saddle point (like the center of a potato chip -- higher in two directions, lower in two others).
Key Concepts
1. Finding Critical Points
Set up the system $f_x(x,y) = 0$ and $f_y(x,y) = 0$ and solve simultaneously. This system may have zero, one, or several solutions.
For polynomial functions, this typically involves solving two polynomial equations simultaneously. Common techniques include substitution or factoring.
2. Saddle Points
A critical point $(a,b)$ where the function has neither a local max nor a local min is a saddle point. The surface looks like a saddle or a mountain pass at that point: if you walk in some directions the value increases; in other directions it decreases. The function $f(x,y) = x^2 - y^2$ has a saddle point at the origin.
Classification requires the second derivative test (next skill).
Worked Example
Find all critical points of $f(x,y) = x^4 + y^4 - 4xy + 1$. (Adapted from Stewart 14.7.)
Setting up the system: $$f_x = 4x^3 - 4y = 0 \implies y = x^3.$$ $$f_y = 4y^3 - 4x = 0 \implies x = y^3.$$
Substituting $y = x^3$ into $x = y^3$: $$x = (x^3)^3 = x^9.$$ $$x^9 - x = 0 \implies x(x^8 - 1) = 0.$$ $$x = 0 \text{ or } x^8 = 1 \implies x = 0, 1, -1.$$
Critical points:
- $x = 0$: $y = 0$. Point $(0,0)$.
- $x = 1$: $y = 1$. Point $(1,1)$.
- $x = -1$: $y = -1$. Point $(-1,-1)$.
To classify these as max, min, or saddle, apply the second derivative test (next skill).
setting $f_x = 0$ alone is sufficient to find potential maxima. A local extremum requires BOTH $f_x = 0$ AND $f_y = 0$. A point where $f_x = 0$ but $f_y \neq 0$ is not a critical point; the function still increases or decreases if you move in the $y$-direction.
Leveled Practice
Problem 1. Find all critical points of $f(x,y) = x^2 + xy + y^2 - 3y$.
Show answer
$f_x = 2x + y = 0 \implies y = -2x$.
$f_y = x + 2y - 3 = 0 \implies x + 2(-2x) = 3 \implies -3x = 3 \implies x = -1$, $y = 2$.
One critical point: $(-1, 2)$.