15.7.Ex27-35

15.7.Ex27-35 - 746 C H A P T E R 15 D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S(ET CHAPTER 14 27 f(x y = 2x y 0 x 1 0 y 3 f is

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746 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 27. f ( x , y ) = 2 x y ,0 x 1 , 0 y 3 SOLUTION f is maximum when x is maximum and y is minimum, that is x = 1and y = 0. f is minimum when x is minimum and y is maximum, that is, x = 0, y = 3. Therefore, the global maximum of f in the set is f ( 1 , 0 ) = 2 · 1 0 = 2 and the global minimum is f ( 0 , 3 ) = 2 · 0 3 =− 3. 28. f ( x , y ) = ( x 2 + y 2 + 1 ) 1 x 3 , 0 y 5 f ( x , y ) = 1 x 2 + y 2 + 1 is maximum when x 2 and y 2 are minimum, that is, when x = y = 0. f is minimum when x 2 and y 2 are maximum, that is, when x = 3and y = 5. Therefore, the global maximum of f on the given set is f ( 0 , 0 ) = ( 0 2 + 0 2 + 1 ) 1 = 1, and the global minimum is f ( 3 , 5 ) = ( 3 2 + 5 2 + 1 ) 1 = 1 35 . 29. f ( x , y ) = e x 2 y 2 , x 2 + y 2 1 The function f ( x , y ) = e ( x 2 + y 2 ) = 1 e x 2 + y 2 is maximum when e x 2 + y 2 is minimum, that is, when x 2 + y 2 is minimum. The minimum value of x 2 + y 2 on the given set is zero, obtained at x = 0and y = 0. We conclude that the maximum value of f on the given set is f ( 0 , 0 ) = e 0 2 0 2 = e 0 = 1 f is minimum when x 2 + y 2 is maximum, that is, when x 2 + y 2 = 1. Thus, the minimum value of f on the given disk is obtained on the boundary of the disk, and it is e 1 = 1 e . 30. Assumptions Matter Show that f ( x , y ) = x + y has no global minimum or maximum on the domain 0 < x , y < 1. Does this contradict Theorem 3? The largest and smallest values of f on the closed square 0 x , y 1are f ( 1 , 1 ) = 2and f ( 1 , 1 ) = 2. However, on the open square 0 < x , y < 1, f can never attain these maximum and minimum values, since the boundary (and in particular the points ( 1 , 1 ) and ( 1 , 1 ) ) are not included in the domain. This does not contradict Theorem 3 since the domain is open. 31. Let D ={ ( x , y ) : x > 0 , y > 0 } . Show that D is not closed. Find a continuous function that does not have a global minimum value on D .
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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15.7.Ex27-35 - 746 C H A P T E R 15 D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S(ET CHAPTER 14 27 f(x y = 2x y 0 x 1 0 y 3 f is

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