Explanation Given: f ( x , y ) = ( x 2 + y 2 ) e − x 2 − y 2 Let P = ( a , b ) be a critical point of f ( x , y ) . Let the second order derivatives f x x , f y y , f x y be continuous near P . Let D be the discriminant of the function f ( x , y ) and is defined as D ( x , y ) = f x x f y y − f x y 2 (a) If D > 0 , then f ( a , b ) is a local minimum or local maximum. (b) If D < 0 , then ( a , b ) is a saddle point. (c) If D = 0 , then the test is inconclusive. (d) If D > 0 and f x x ( a , b ) > 0 then f ( a , b ) is a local minimum. (a) If D > 0 and f x x ( a , b ) < 0 then f ( a , b ) is a local maximum. Calculation: (a) It can be observed that the exponential term e − x 2 − y 2 cannot be zero or negative. The polynomial term ( x 2 + y 2 ) can also be neither zero nor negative. Hence, the minimum value of the function f can be zero at ( 0 , 0 ) (b) f ( x , y ) = ( x 2 + y 2 ) e − x 2 − y 2 f x ( x , y ) = ( x 2 + y 2 ) ( e − x 2 − y 2 ) ( − 2 x ) + 2 x e − x 2 − y 2 = e − x 2 − y 2 ( − 2 x 3 − 2 x y 2 + 2 x ) f x ( x , y ) = 0 ⇒ e − x 2 − y 2 ( − 2 x 3 − 2 x y 2 + 2 x ) = 0 − 2 x 3 − 2 x y 2 + 2 x = 0 ( e − x 2 − y 2 ≠ 0 ) 2 x ( 1 − y 2 − x 2 ) = 0 x ( 1 − y 2 − x 2 ) = 0 x = 0 , 1 − y 2 − x 2 = 0 x = 0 , x 2 + y 2 = 1 f y

4th Edition

ISBN: 9781319308858

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 14.7, Problem 26E

To determine

(a) Where the function $f$ takes on its minimum value without using the calculus.

(b) That the set of critical points of $f$ consists of the origin and the unit circle $x^{2} + y^{2} = 1$

(c) That $f$ takes the maximum value on the unit circle by analyzing the function $g (t) = t e^{- t} (t > 0)$

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Chapter 14.7, Problem 25E

Chapter 14.7, Problem 27E

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Chapter 14 Solutions

CALCULUS:EARLY TRANS.(LL)-W/ACCESS

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

Knowledge Booster

(a) Where the function f takes on its minimum value without using the calculus. (b) That the set of critical points of f consists of the origin and the unit circle x 2 + y 2 = 1 (c) That f takes the maximum value on the unit circle by analyzing the function g ( t ) = t e − t ( t &gt; 0 )

(a) Where the function f takes on its minimum value without using the calculus. (b) That the set of critical points of f consists of the origin and the unit circle x2+y2=1 (c) That f takes the maximum value on the unit circle by analyzing the function g(t)=te−t(t>0)

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Chapter 14 Solutions

(a) Where the function f takes on its minimum value without using the calculus. (b) That the set of critical points of f consists of the origin and the unit circle x 2 + y 2 = 1 (c) That f takes the maximum value on the unit circle by analyzing the function g ( t ) = t e − t ( t > 0 )

(a) Where the function $f$ takes on its minimum value without using the calculus.

(b) That the set of critical points of $f$ consists of the origin and the unit circle $x^{2} + y^{2} = 1$

(c) That $f$ takes the maximum value on the unit circle by analyzing the function $g (t) = t e^{- t} (t > 0)$