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15.7 Ex 41%2 - 752 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 By Theorem 3 the extreme values

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752 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) By Theorem 3, the extreme values occur either at a critical point in the interior of the square or at a point on the boundary of the square. Since there are no critical points in the interior of the square, the candidates for extreme values are the following points: ( 0 , 0 ), ( 1 , 0 ), ( 1 , 1 ), ( 0 , 1 ) We compute f ( x , y ) = x 3 + y 3 3 xy at these points: f ( 0 , 0 ) = 0 3 + 0 3 3 · 0 = 0 f ( 1 , 0 ) = 1 3 + 0 3 3 · 1 · 0 = 1 f ( 1 , 1 ) = 1 3 + 1 3 3 · 1 · 1 =− 1 f ( 0 , 1 ) = 0 3 + 1 3 3 · 0 · 1 = 1 We conclude that in the given domain, the global maximum is f ( 1 , 0 ) = f ( 0 , 1 ) = 1 and the global minimum is f ( 1 , 1 ) 1. 41. f ( x , y ) = x 3 y 5 , the set bounded by x = 0, y = 0, and y = 1 x . Hint: Use a computer algebra system to Fnd the minimum along the boundary curve y = 1 x , which is parametrized by ( t , 1 t ) for 0 t 1. SOLUTION x y = 1 x y A (1, 0) 0 B (0, 1) We use the following steps. Step 1. Examine the critical points. We Fnd the critical points of f ( x , y ) = x 3 y 5 in the interior of the domain. Setting the partial derivatives equal to zero and solving gives f x ( x , y ) = 3 x 2 y 5 = 0 , f y ( x , y ) = 5 y 4 x 3 = 0 The solutions are all the points with at least one zero coordinate. These points are not in the interior of the region, hence there are no critical points in the interior of the region. Step 2. Check the boundary. We consider each part of the boundary separately. The segment OA : On this segment y = 0and f ( x , 0 ) = 0, hence f has the constant value 0 on this part of the boundary. The segment OB : On this segment x = 0, hence f ( 0 , y ) = 0 3 · y 5 = 0. f attains the constant value 0 on this part of the boundary. The curve y = 1 x ,0 x 1: On this curve we have f ( x , y ) = g ( x ) = x 3 ( 1 x ) 5 . We Fnd the points on the interval 0 x 1where g ( x ) = x 3 ( 1 x ) 5 has maximum and minimum values. Differen- tiating g gives g ± ( x ) = 3 x 2 ( 1 x ) 5 + x 3 · 5 ( 1 x ) 4 · ± 1 2 x = 3 x 2 ( 1 x ) 5 5 2 x 2 . 5 ( 1 x ) 4 = x 2 ( 1 x ) 4 ± 3 3 x 5 2 x = x 2 ( 1 x ) 4 ± 3 11 2 x We solve g ± ( x ) = 0 in the interval 0 < x < 1: g ± ( x ) = x 2 ( 1 x ) 4 ± 3 11 2 x = 0 x = 0 , 1 x = 0 , 3 11 2 x = 0 The solutions are x = 0, x = 1, and x = 36 121 . The critical point in the interval 0 < x < 1is x = 36 121 . We compute g ( x ) = x 3 ( 1 x ) 5 at the critical point and at the endpoints x = 0, x = 1: g ( 0 ) = g ( 1 ) = 0 , g ± 36 121 = ± 36 121 3 ± 1 6 11 5 0 . 0005
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SECTION 15.7 Optimization in Several Variables (ET Section 14.7) 753 The points where g has extreme values in the interval 0 x 1a re x = 0, x = 1, and x = 36 121 .W efndthe
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.7 Ex 41%2 - 752 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 By Theorem 3 the extreme values

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