15.8.Ex20-21

15.8.Ex20-21 - 780 C H A P T E R 15 D I F F E R E N T I AT...

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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 à a + b 2 a !
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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.Ex20-21 - 780 C H A P T E R 15 D I F F E R E N T I AT...

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