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• We have fæe = 12x? – 16, fyy = 6, fay = 0. - Hence frzfyy - fry = (-16)(6) < 0 at (x, y) = (0, 1) so f(x, y) has a saddle point at (0, 1), and fra fyy – fay = (32)(6) > 0, at (x, y) = (±2, 1). - YY Since fyy > 0, f(x,y) has minima at (±2, 1).

• We have fæe = 12x? – 16, fyy = 6, fay = 0. - Hence frzfyy - fry = (-16)(6) < 0 at (x, y) = (0, 1) so f(x, y) has a saddle point at (0, 1), and fra fyy – fay = (32)(6) > 0, at (x, y) = (±2, 1). - YY Since fyy > 0, f(x,y) has minima at (±2, 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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10. Find all critical points of the function f (x, y) = x* – 8x² + 3y? – 6y. and classify them as maximums, minimums, or saddle-point. Solution. • At a critical point fr = 4x° – 16x = = 0, fy = 6y – 6 = 0. If follows that y = 1 and x = 0,±2 so the critical points of f are (п, у) — (0, 1), (2, 1), (-2, 1). • We have fae = 12x? – 16, fyy = 6, fry = 0. Hence fra fyy – fay = (-16)(6) < 0 so f(x, y) has a saddle point at (0,1), and at (x, y) = (0, 1) faz fyy – fay = (32)(6) > 0, at (x, y) = (±2, 1). Since fyy > 0, f (x, y) has minima at (+2, 1).
Transcribed Image Text:10. Find all critical points of the function f (x, y) = x* – 8x² + 3y? – 6y. and classify them as maximums, minimums, or saddle-point. Solution. • At a critical point fr = 4x° – 16x = = 0, fy = 6y – 6 = 0. If follows that y = 1 and x = 0,±2 so the critical points of f are (п, у) — (0, 1), (2, 1), (-2, 1). • We have fae = 12x? – 16, fyy = 6, fry = 0. Hence fra fyy – fay = (-16)(6) < 0 so f(x, y) has a saddle point at (0,1), and at (x, y) = (0, 1) faz fyy – fay = (32)(6) > 0, at (x, y) = (±2, 1). Since fyy > 0, f (x, y) has minima at (+2, 1).
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