15.8.Ex1-6 - S E C T I O N 15.8 Lagrange Multipliers:...

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SECTION 15.8 Lagrange Multipliers: Optimizing with a Constraint (ET Section 14.8) 765 2 6 2 26 g ( x , y ) = 0 f A , g A A E C D B 2 6 6 (b) The minimum and maximum occur where the level curve of f is tangent to the constraint curve. The level curves tangent to the constraint curve are f ( A ) =− 4 , f ( C ) = 2 , f ( B ) = 6 , f ( D ) 4 , f ( E ) = 4 Therefore the global minimum of f under the constraint is 4 and the global maximum is 6. Exercises 1. Use Lagrange multipliers to Fnd the extreme values of the function f ( x , y ) = 2 x + 4 y subject to the constraint g ( x , y ) = x 2 + y 2 5 = 0. (a) Show that the Lagrange equation f = λ g gives x = 1and y = 2. (b) Show that these equations imply 6= 0and y = 2 x . (c) Use the constraint equation to determine the possible critical points ( x , y ) . (d) Evaluate f ( x , y ) at the critical points and determine the minimum and maximum values. SOLUTION (a) The Lagrange equations are determined by the equality f = g .WeFndthem: f = - f x , f y ® = h 2 , 4 i , g = - g x , g y ® = h 2 x , 2 y i Hence, h 2 , 4 i = h 2 x , 2 y i or ( 2 x ) = 2 ( 2 y ) = 4 x = 1 y = 2 (b) The Lagrange equations in part (a) imply that 0. The Frst equation implies that x = 1 and the second equation gives y = 2 . Therefore y = 2 x . (c) We substitute y = 2 x in the constraint equation x 2 + y 2 5 = 0 and solve for x and y .Thisgives x 2 + ( 2 x ) 2 5 = 0 5 x 2 = 5 x 2 = 1 x 1 1 , x 2 = 1 Since y = 2 x ,wehave y 1 = 2 x 1 2, y 2 = 2 x 2 = 2. The critical points are thus ( 1 , 2 ) and ( 1 , 2 ). Extreme values can also occur at the points where g = h 2 x , 2 y i = h 0 , 0 i .However , ( 0 , 0 ) is not on the constraint. (d) We evaluate f ( x , y ) = 2 x + 4 y at the critical points, obtaining f ( 1 , 2 ) = 2 · ( 1 ) + 4 · ( 2 ) 10 f ( 1 , 2 ) = 2 · 1 + 4 · 2 = 10 Since f is continuous and the graph of g = 0 is closed and bounded, global minimum and maximum points exist. So according to Theorem 1, we conclude that the maximum of f ( x , y ) on the constraint is 10 and the minimum is 10. 2. ±ind the extreme values of f ( x , y ) = x 2 + 2 y 2 subject to the constraint g ( x , y ) = 4 x 6 y = 25. (a) Show that the Lagrange equations yield 2 x = 4 ,4 y 6 . (b) Show that if x = 0or y = 0, then = 0 and the Lagrange equations give x = y = 0. Since ( 0 , 0 ) does not satisfy the constraint, you may assume that x and y are nonzero.
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766 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) (c) Use the Lagrange equations to show that y =− 3 4 x . (d) Substitute in the constraint equation to show that there is a unique critical point P . (e) Does P correspond to a minimum or maximum value of f ? Refer to Figure 10 to justify your answer. Hint: Do the values of f ( x , y ) increase or decrease as ( x , y ) moves away from P along the line g ( x , y ) = 0? 4 6 2 0 4 2 8 02 6 4 6 12 24 36 g ( x , y ) = 0 6 8 4 2 P FIGURE 10 Level curves of f ( x , y ) = x 2 + 2 y 2 and constraint g ( x , y ) = 4 x 6 y 25 = 0. SOLUTION (a) The gradients f and g are f = h 2 x , 4 y i , g = h 4 , 6 i The Lagrange equations are thus f = λ g h 2 x , 4 y i = h 4 , 6 i or 2 x = 4 4 y 6 (b) If x = 0, the ±rst equation gives 0 = 4 or = 0. Substituting in the second equation gives 4 y = 0or y = 0.
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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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15.8.Ex1-6 - S E C T I O N 15.8 Lagrange Multipliers:...

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