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15.7.Ex1-5

# 15.7.Ex1-5 - S E C T I O N 15.7 Optimization in Several...

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S E C T I O N 15.7 Optimization in Several Variables (ET Section 14.7) 729 Point S in Figure (B) is neither a local extremum nor a saddle point of f . 1 3 0 1 3 S Figure (B) 3. Let f ( x , y ) be a continuous function on a domain D in R 2 . Determine which of the following statements are true: (a) If D is closed and bounded, then f takes on a maximum value on D . (b) If D is neither closed nor bounded, then f does not take on a maximum value of D . (c) f ( x , y ) need not have a maximum value on D = { ( x , y ) : 0 x , y 1 } . (d) A continuous function takes on neither a minimum nor a maximum value on the open quadrant { ( x , y ) : x > 0 , y > 0 } . SOLUTION (a) This statement is true. It follows by the Theorem on Existence of Global Extrema. (b) The statement is false. Consider the constant function f ( x , y ) = 2 in the following domain: x 1 y D = { ( x , y ) : 0 < x 1 , 0 y < ∞} Obviously f is continuous and D is neither closed nor bounded. However, f takes on a maximum value (which is 2) on D . (c) The domain D = { ( x , y ) : 0 x , y 1 } is the following rectangle: x 1 y 1 D = { ( x , y ) : 0 x , y 1 } D is closed and bounded, hence f takes on a maximum value on D . Thus the statement is false. (d) The statement is false. The constant function f ( x , y ) = c takes on minimum and maximum values on the open quadrant. Exercises 1. Let P = ( a , b ) be a critical point of f ( x , y ) = x 2 + y 4 4 xy . (a) First use f x (, x , y ) = 0 to show that a = 2 b . Then use f y ( x , y ) = 0 to show that P = ( 0 , 0 ) , ( 2 2 , 2 ) , or ( 2 2 , 2 ) . (b) Referring to Figure 15, determine the local minima and saddle points of f ( x , y ) and find the absolute minimum value of f ( x , y ) .

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730 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) x y z FIGURE 15 SOLUTION (a) We find the partial derivatives: f x ( x , y ) = x x 2 + y 4 4 xy = 2 x 4 y f y ( x , y ) = y x 2 + y 4 4 xy = 4 y 3 4 x Since P = ( a , b ) is a critical point, f x ( a , b ) = 0. That is, 2 a 4 b = 0 a = 2 b Also f y ( a , b ) = 0, hence, 4 b 3 4 a = 0 a = b 3 We obtain the following equations for the critical points ( a , b ) : a = 2 b a = b 3 Equating the two equations, we get 2 b = b 3 b 3 2 b = b ( b 2 2 ) = 0 b 1 = 0 b 2 = 2 b 3 = − 2 Since a = 2 b , we have a 1 = 0, a 2 = 2 2, a
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15.7.Ex1-5 - S E C T I O N 15.7 Optimization in Several...

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