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...

This preview shows pages 1–2. Sign up to view the full content.

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 !

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online