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11. Let {(x, y) : x² + y² < 1} D = be the unit disc and f(x, y) = x² – 2x + y² + 2y + 1. Find the global maximum and minimum of f : D → R At what points (x, y) in D does f attain its maximum and minimum? Solution. • The function is differentiable everywhere. A critical point (x, y) satisfies fa = 2x – 2 = 0, 2y + 2 = 0 which implies that (x, y) = (1,-1). This point does not lie inside D, so f has no critical points inside D and it must attain its global maximum and minimum on the boundary of D. • On the boundary x² + y² = 1, we can write cos 0, y = sin 0 where 0 < 0 < 2T is the polar angle of (x, y). Then f(cos 0, sin 0) = cos² 0 – 2 cos 0+sin? 0+2 sin 0 +1 = 2– 2 cos 0+2 sin 0.

11. Let {(x, y) : x² + y² < 1} D = be the unit disc and f(x, y) = x² – 2x + y² + 2y + 1. Find the global maximum and minimum of f : D → R At what points (x, y) in D does f attain its maximum and minimum? Solution. • The function is differentiable everywhere. A critical point (x, y) satisfies fa = 2x – 2 = 0, 2y + 2 = 0 which implies that (x, y) = (1,-1). This point does not lie inside D, so f has no critical points inside D and it must attain its global maximum and minimum on the boundary of D. • On the boundary x² + y² = 1, we can write cos 0, y = sin 0 where 0 < 0 < 2T is the polar angle of (x, y). Then f(cos 0, sin 0) = cos² 0 – 2 cos 0+sin? 0+2 sin 0 +1 = 2– 2 cos 0+2 sin 0.

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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11. Let D = {(x, y) : a² + y < 1} be the unit disc and f (x, y) = x² – 2x + y + 2y + 1. Find the global maximum and minimum of f : D → R At what points (x, y) in D does f attain its maximum and minimum? Solution. • The function is differentiable everywhere. A critical point (x, y) satisfies fa = 2x – 2 = 0, 2y +2 = 0 which implies that (x, y) = (1, -1). This point does not lie inside D, so f has no critical points inside D and it must attain its global maximum and minimum on the boundary of D. • On the boundary x? + y? = 1, we can write cos 0, y = sin 0 X = where 0 < 0 < 2n is the polar angle of (x, y). Then f (cos 0, sin 0) = cos² 0 – 2 cos 0+sin? 0+2 sin 0+1=2-2 cos 0+2 sin 0. • Since f(cos 0, sin 0) is a differentiable function of 0, the maximum and minimum of ƒ on the boundary are attained at a critical point where d of (cos 0, sin 0) = 2 sin 0 + 2 cos 0 = 0. It follows that tan 0 = -1, so 0 = 37/4 or 0 =
Transcribed Image Text:11. Let D = {(x, y) : a² + y < 1} be the unit disc and f (x, y) = x² – 2x + y + 2y + 1. Find the global maximum and minimum of f : D → R At what points (x, y) in D does f attain its maximum and minimum? Solution. • The function is differentiable everywhere. A critical point (x, y) satisfies fa = 2x – 2 = 0, 2y +2 = 0 which implies that (x, y) = (1, -1). This point does not lie inside D, so f has no critical points inside D and it must attain its global maximum and minimum on the boundary of D. • On the boundary x? + y? = 1, we can write cos 0, y = sin 0 X = where 0 < 0 < 2n is the polar angle of (x, y). Then f (cos 0, sin 0) = cos² 0 – 2 cos 0+sin? 0+2 sin 0+1=2-2 cos 0+2 sin 0. • Since f(cos 0, sin 0) is a differentiable function of 0, the maximum and minimum of ƒ on the boundary are attained at a critical point where d of (cos 0, sin 0) = 2 sin 0 + 2 cos 0 = 0. It follows that tan 0 = -1, so 0 = 37/4 or 0 =
• The second derivative of f along the boundary is af(cos 0, sin 0) = 2 cos 0 – 2 sin 0 d20 We have 1 1 1 1 sin 4 sin COS 4 V2 V2 COS 4 V2 SO df < 0 d20 at 0 = 37/4, > 0 d²0 at 0 77/4. Thus, f has a maximum value at 0 0 = 77/4. 37/4 and a minimum value at • Evaluating the corresponding values of (x, y) and f, we find that f has a global maximum on D at 1 1 (x, y) where f(x, y) = 2+ 2v2, V2' V2, and a global minimum on D at 1 (x, y) where f(x, y) = 2 – 2/2. V2' V2,
Transcribed Image Text:• The second derivative of f along the boundary is af(cos 0, sin 0) = 2 cos 0 – 2 sin 0 d20 We have 1 1 1 1 sin 4 sin COS 4 V2 V2 COS 4 V2 SO df < 0 d20 at 0 = 37/4, > 0 d²0 at 0 77/4. Thus, f has a maximum value at 0 0 = 77/4. 37/4 and a minimum value at • Evaluating the corresponding values of (x, y) and f, we find that f has a global maximum on D at 1 1 (x, y) where f(x, y) = 2+ 2v2, V2' V2, and a global minimum on D at 1 (x, y) where f(x, y) = 2 – 2/2. V2' V2,
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