Optimization in two variables and Lagrange multipliers practice question 6
How many of the critical points of are saddle points?
splits into a part in alone and a part in alone. How does that help you find every critical point, and what then tells a saddle apart from a maximum or a minimum?
and . Each involves one variable only, so every root of one pairs with every root of the other.
, and . Find the sign of at each critical point.
A whole number.
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
Both partials must vanish, and each depends on one variable only.
Every x pairs with every y: four critical points.
With no mixed term, D is the product of the two pure second partials.
(1, 2) has f_xx = 6 > 0, a local minimum; (−1, 0) has f_xx = −6 < 0, a local maximum.
The two points where D is negative.
Answer
The answer is 2, the saddle points (1, 0) 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.
The idea behind it
The optimization in two variables and Lagrange multipliers lesson, a figure to move, in a few steps
Leans on
Second partial derivatives, from Functions of two variables and partial derivatives
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