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15.8%20Ex%2020%20-%2028

# 15.8%20Ex%2020%20-%2028 - 780 C H A P T E R 15 D I F F E R...

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780 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 20. Show that the point ( x 0 , y 0 ) closest to the origin on the line ax + by = c has coordinates x 0 = ac a 2 + b 2 , y 0 = bc a 2 + b 2 SOLUTION We need to minimize the distance d ( x , y ) = x 2 + y 2 subject to the constraint g ( x , y ) = ax + by = c . Notice that the distance d ( x , y ) is at a minimum at the same points where the square of the distance d 2 ( x , y ) is at a minimum (since the function u 2 is increasing for u 0). Therefore, we may find the minimum of f ( x , y ) = x 2 + y 2 subject to the constraint ax + by = c . Step 1. Write out the Lagrange Equations. The gradient vectors are f = 2 x , 2 y and g = a , b , hence the Lagrange Condition f = λ g is 2 x , 2 y = λ a , b or 2 x = λ a 2 y = λ b Step 2. Solve for λ in terms of x and y . The Lagrange equations give λ = 2 x a and λ = 2 y b Step 3. Solve for x and y using the constraint. We equate the two expressions for λ and solve for y in terms of x : 2 x a = 2 y b y = b a x We now substitute y = bx a in the equation of the constraint ax + by = c and solve for x : ax + b · b a x = c a + b 2 a x = c a 2 + b 2 a x = c x = ac a 2 + b 2 We find y using the relation y = bx a : y = b a · ac a 2 + b 2 = bc a 2 + b 2 The critical point is thus x 0 = ac a 2 + b 2 , y 0 = bc a 2 + b 2 (1) Step 4. Conclusions. It is clear geometrically that the problem has a minimum value and it does not have a maximum value. Therefore the minimum occurs at the critical point. We conclude that the point closest to the origin on the line ax + by = c is given by (1). To show that the vector x 0 , y 0 is perpendicular to the line, we write the line in vector form as x x 0 , y y 0 · a , b = 0. Thus, a , b is perpendicular to the line. Since x 0 , y 0 = c a 2 + b 2 a , b , then x 0 , y 0 is parallel to a , b , and thus also perpendicular to the line. 21. Find the maximum value of f ( x , y ) = x a y b for x , y 0 on the unit circle, where a , b > 0 are constants. SOLUTION We must find the maximum value of f ( x , y ) = x a y b ( a , b > 0) subject to the constraint g ( x , y ) = x 2 + y 2 = 1. Step 1. Write out the Lagrange Equations. We have f = ax a 1 y b , bx a y b 1 and g = 2 x , 2 y . Therefore the Lagrange Condition f = λ g is ax a 1 y b , bx a y b 1 = λ 2 x , 2 y or ax a 1 y b = 2 λ x bx a y b 1 = 2 λ y (1)

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S E C T I O N 15.8 Lagrange Multipliers: Optimizing with a Constraint (ET Section 14.8) 781 Step 2. Solve for λ in terms of x and y . If x = 0 or y = 0, f has the minimum value 0. We thus may assume that x > 0 and y > 0. The equations (1) imply that λ = ax a 2 y b 2 , λ = bx a y b 2 2 Step 3. Solve for x and y using the constraint. Equating the two expressions for λ and solving for y in terms of x gives ax a 2 y b 2 = bx a y b 2 2 ax a 2 y b = bx a y b 2 ay 2 = bx 2 y 2 = b a x 2 y = b a x We now substitute y = b a x in the constraint x 2 + y 2 = 1 and solve for x > 0. We obtain x 2 + b a x 2 = 1 ( a + b ) x 2 = a x 2 = a a + b x = a a + b We find y using the relation y = b a x : y = b a a a + b = ab a ( a + b ) = b a + b We obtain the critical point: a a + b , b a + b Extreme points can also occur where g = 0 , that is, 2 x , 2 y = 0 , 0 or ( x , y ) = ( 0 , 0 ) . However, this point is not on the constraint.
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15.8%20Ex%2020%20-%2028 - 780 C H A P T E R 15 D I F F E R...

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