Explanation Given: Let P = ( a , b ) be the critical point of f ( x , y ) = x 2 + y 4 − 4 x y 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 are 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. (e) If D > 0 and f x x ( a , b ) < 0 , then f ( a , b ) is a local maximum. Calculation: (a) f ( x , y ) = x 2 + y 4 − 4 x y Differentiating we get, f x = 2 x − 4 y f y = 4 y 3 − 4 x Since P = ( a , b ) be the critical point, so substituting x = a , y = b in f x ( x , y ) = 0 2 x − 4 y = 0 2 a − 4 b = 0 a = 2 b Hence, the critical point P = ( a , b ) satisfies a = 2 b Computing f x = 0 , f y = 0 we get 2 x − 4 y = 0... ( 1 ) 4 y 3 − 4 x = 0... ( 2 ) Putting 4 x = 8 y from ( 1 ) into ( 2 ) 4 y 3 − 8 y = 0 4 y ( y 2 − 2 ) = 0 y = 0 , y = ± 2 If y = 0 , then x = 0 If y = 2 , then x = 2 2 If y = − 2 , then x = 2 2 Hence, the critical points are P = ( 0 , 0 ) ( 2 2 , 2 ) , ( − 2 2 , − 2 ) (b) Here, we have f x x ( x , y ) = 2 f y y ( x , y ) = 12 y 2 f x y ( x , y ) = − 4 Now, D ( x , y ) = f x x ( x , y ) f y y ( x , y ) − f x y ( x , y ) 2 = 2...

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

4th Edition

ISBN: 9781319057664

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 14.7, Problem 1E

To determine

(a) Show that $a = 2 b$ using $f_{x} (x, y) = 0$ . And, show that $P = (0, 0) (2 \sqrt{2,} \sqrt{2}), (- 2 \sqrt{2}, - \sqrt{2})$ using $f_{y} (x, y) = 0$

(b) The local minima and saddle points of $f (x, y)$ . Also,the absolute minimum of $f (x, y)$ .

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Chapter 14.7, Problem 3PQ

Chapter 14.7, Problem 2E

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Chapter 14 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

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) Show that a = 2 b using f x ( x , y ) = 0 . And, show that P = ( 0 , 0 ) ( 2 2 , 2 ) , ( − 2 2 , − 2 ) using f y ( x , y ) = 0 (b) The local minima and saddle points of f ( x , y ) . Also,the absolute minimum of f ( x , y ) .

(a) Show that a=2b using fx(x,y)=0. And, show that P=(0,0)(22,2),(−22,−2) using fy(x,y)=0 (b) The local minima and saddle points of f(x,y). Also,the absolute minimum of f(x,y).

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Chapter 14 Solutions

(a) Show that $a = 2 b$ using $f_{x} (x, y) = 0$ . And, show that $P = (0, 0) (2 \sqrt{2,} \sqrt{2}), (- 2 \sqrt{2}, - \sqrt{2})$ using $f_{y} (x, y) = 0$

(b) The local minima and saddle points of $f (x, y)$ . Also,the absolute minimum of $f (x, y)$ .