Optimization in two variables and Lagrange multipliers, question 5

Topic: Optimization in two variables and Lagrange multipliers
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §16.1Key results

How many of the critical points of $f(x, y) = x^{3} - 3x + y^{3} - 3y^{2}$ are saddle points?

Your answer

A whole number.

How sure are you?

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Theory UBC Math 100 §16.1Key results

Worked solution

Strategy

Find every critical point, then classify each with D = f_xx f_yy − (f_xy)². A critical point is a saddle exactly when D < 0.

Working

$f_x = 3x^{2} - 3 = 0 \ \Rightarrow\ x = \pm 1; \qquad f_y = 3y(y - 2) = 0 \ \Rightarrow\ y = 0 \text{ or } 2$
Both partials must vanish, and each depends on one variable only.
$(1, 0), \quad (1, 2), \quad (-1, 0), \quad (-1, 2)$
Every x pairs with every y: four critical points.
$f_{xx} = 6x, \quad f_{yy} = 6y - 6, \quad f_{xy} = 0 \qquad\Longrightarrow\qquad D = 36x(y - 1)$
With no mixed term, D is the product of the two pure second partials.
$D(1, 0) = -36, \quad D(1, 2) = 36, \quad D(-1, 0) = 36, \quad D(-1, 2) = -36$
(1, 2) has f_xx = 6 > 0, a local minimum; (−1, 0) has f_xx = −6 < 0, a local maximum.
$\text{saddle points: } (1, 0) \text{ and } (-1, 2)$
The two points where D is negative.

Answer

2 \ \text{saddle points, at } (1, 0) \text{ and } (-1, 2)

Check

The function splits as (x³ − 3x) + (y³ − 3y²), each part with one local maximum and one local minimum, and a saddle pairs a maximum of one with a minimum of the other: 2 ways.

Where this goes wrong

Classifying each point by the sign of f_xx alone, which calls every point a maximum or a minimum and finds no saddles. Find the sign of D first; f_xx only matters when D > 0.