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 CHAPTER 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 ) = p x 2 + y 2 subject to the constraint g ( x , y ) = + = 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 Fnd the minimum of f ( x , y ) = x 2 + y 2 subject to the constraint + = c . Step 1. Write out the Lagrange Equations. The gradient vectors are f = h 2 x , 2 y i and g = h a , b i , hence the Lagrange Condition f = λ g is h 2 x , 2 y i = h a , b i or 2 x = a 2 y = b Step 2. Solve for in terms of x and y . The Lagrange equations give = 2 x a and = 2 y b Step 3. Solve for x and y using the constraint. We equate the two expressions for 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 + = c and solve for x : + b · b a x = c Ã a + b 2 a ! x = c a 2 + b 2 a x = c x = a 2 + b 2 We Fnd y using the relation y = a : y = b a · a 2 + b 2 = bc a 2 + b 2 The critical point is thus x 0 = 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 + = c is given by (1). To show that the vector h x 0 , y 0 i is perpendicular to the line, we write the line in vector form as h x x 0 , y y 0 i·h a , b i= 0. Thus, h a , b i is perpendicular to the line. Since h x 0 , y 0 c a 2 + b 2 h a , b i ,then h x 0 , y 0 i is parallel to h a , b i , and thus also perpendicular to the line. 21. ±ind 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. We must Fnd 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 = D a 1 y b , a y b 1 E and g = h 2 x , 2 y i . Therefore the Lagrange Condition f = g is D a 1 y b , a y b 1 E = h 2 x , 2 y i or a 1 y b = 2 x a y b 1 = 2 y (1)

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SECTION 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 = 0or y = 0, f has the minimum value 0. We thus may assume that x > 0 and y > 0. The equations (1) imply that = ax a 2 y b 2 , = bx a y b 2 2 Step 3. Solve for x and y using the constraint. Equating the two expressions for and solving for y in terms of x gives a 2 y b 2 = a y b 2 2 a 2 y b = a y b 2 ay 2 = 2 y 2 = b a x 2 y = ± b a x We now substitute y = q 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 Fnd y using the relation y = q b a x : y = ± b a ± a a + b = s ab a ( a + b ) = ± b a + b We obtain the critical point: Ã ± a a + b , ± b a + b !
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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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15.8%20Ex%2020%20-%2028 - 780 C H A P T E R 15 D I F F E R...

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