Questions: Critical Points of Multivariable Functions

Question 1

Find all critical points of f(x, y) = x³ − 3x + y² − 4y.

Model answer: Set f_x = 3x² − 3 = 0, giving x = ±1. Set f_y = 2y − 4 = 0, giving y = 2. Critical points are (1, 2) and (−1, 2).

Each partial derivative is set to zero independently. f_x = 0 gives x² = 1, so x = 1 or x = −1. f_y = 0 gives y = 2. These combine to yield two critical points. Both are candidates for extrema; the second derivatives test distinguishes their nature.

Question 2

A critical point is generally a local maximum or local minimum.

• T. True
• F. False

Answer: False

Saddle points are critical points that are neither local maxima nor local minima. At a saddle point, the function increases in some directions and decreases in others. The classic example is f(x, y) = x² − y², whose critical point at the origin is a saddle.

Question 3

Why must you also check boundary points when optimizing a function on a closed bounded domain, not just interior critical points?

Model answer: On a closed bounded domain, a continuous function is guaranteed (by the extreme value theorem) to attain its global maximum and minimum somewhere on the domain. These extrema may occur either at interior critical points (where ∇f = 0) or on the boundary. Interior critical points are only candidates for extrema inside the open domain; the boundary is a separate region that must be analyzed independently.

Consider f(x, y) = x + y on the unit disk x² + y² ≤ 1. The only critical point of the unrestricted function is nowhere (∇f = (1,1) ≠ 0 everywhere), so all extrema are on the boundary. The maximum is at (1/√2, 1/√2) and the minimum at (−1/√2, −1/√2), both on the boundary circle.