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EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + yA - 4xy + 1.

EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + yA - 4xy + 1.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + yA – 4xy + 1. SOLUTION We first locate the critical points: fy = 4y3 – 4x. Setting these partial derivatives equal to 0, we obtain the equations Video Example = 0 and y3 - x = 0. To solve these equations we substitute y = x³ from the first equation into the second one. This gives 0 = = x9 - x = x(x8 - 1) – x + 1) = + 1)(x* + 1) so there are three real roots: x = 0, 1, The three critical points are (0, 0), (1, 1), and (-1, -1). Next we calculate the second partial derivatives and D(x, y): fxx fxy fyy fxfyy - (fxy)2 = D(x, y) Since D(0, 0) = -16 < 0, it follows from the Second Derivative Test that the origin is a saddle point; that is, f has no local maximum or minimum at (0, 0). Since D(1, 1) = > 0 and fy(1, 1) = 12 > 0, we see that f(1, 1) = -1 is a local ( ---Select-- :. Similarly, we have D(-1, –1) = > 0 and fyx(-1, -1) = 12 > 0, so f(-1, -1) = -1 is also a local ---Select---
Transcribed Image Text:EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + yA – 4xy + 1. SOLUTION We first locate the critical points: fy = 4y3 – 4x. Setting these partial derivatives equal to 0, we obtain the equations Video Example = 0 and y3 - x = 0. To solve these equations we substitute y = x³ from the first equation into the second one. This gives 0 = = x9 - x = x(x8 - 1) – x + 1) = + 1)(x* + 1) so there are three real roots: x = 0, 1, The three critical points are (0, 0), (1, 1), and (-1, -1). Next we calculate the second partial derivatives and D(x, y): fxx fxy fyy fxfyy - (fxy)2 = D(x, y) Since D(0, 0) = -16 < 0, it follows from the Second Derivative Test that the origin is a saddle point; that is, f has no local maximum or minimum at (0, 0). Since D(1, 1) = > 0 and fy(1, 1) = 12 > 0, we see that f(1, 1) = -1 is a local ( ---Select-- :. Similarly, we have D(-1, –1) = > 0 and fyx(-1, -1) = 12 > 0, so f(-1, -1) = -1 is also a local ---Select---
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