# HW06 - HW 6 Solutions Math 1920 Spring 2009 March 3 2009...

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Unformatted text preview: HW 6 Solutions Math 1920 Spring 2009 March 3, 2009 Problem 14.8.1 Let g ( x,y ) = x 2 + 2 y 2- 1 so that our constraint is g = 0. Then ∇ f = y i + x j and ∇ g = 2 x i + 4 y j . Setting ∇ f = λ ∇ g gives the equations y = 2 xλ and x = 4 yλ . Substituting we get x = 8 xλ 2 , so either x = 0 or λ = ± √ 2 4 . If x = 0, then y = 0, but (0,0) is not on the ellipse, so we can not have x = 0. Therefore λ = ± √ 2 4 , so x = ± √ 2 y . Substituting this into g ( x,y ) = 0 gives 2 y 2 + 2 y 2- 1 = 0, so y = ± 1 2 So the extreme values of f on the ellipse occur at ± √ 2 2 , 1 2 and ± √ 2 2 ,- 1 2 . The extreme values of f on the ellipse are ± √ 2 4 . Problem 14.8.13 Let f ( x,y ) = x 2 + y 2 and g ( x,y ) = x 2- 2 x + y 2- 4 y . Then ∇ f = 2 x i + 2 y j and ∇ g = (2 x- 2) i + (2 y- 4) j . Setting ∇ f = λ ∇ g gives the equations 2 x = (2 x- 2) λ and 2 y = (2 y- 4) λ . Solving we get x = 1 λ- 1 and y = 2 λ- 1 for λ 6 = 1. If λ = 1 we have 2 x = 2 x- 2, which is impossible. Therefore y = 2 x . Substituting this into g ( x,y ) = 0 gives 0 = x 2- 2 x + 4 x 2- 8 x = 5 x 2- 10 x = 5 x ( x- 2). So either x = 0, in which case y = 0, or x = 2, in which case y = 4. The minimum value of f on the circle is f (0 , 0) = 0 and the maximum value is f (2 , 4) = 20. Problem 14.8.17 We will minimize the square of the distance from the points on the plane to (1,1,1). The square the distance to (1,1,1) is given by the function f ( x,y,z ) = ( x- 1) 2 +( y- 1) 2 +( z- 1) 2 . Let g ( x,y,z ) = x +2 y +3 z- 13. Then ∇ f = (2 x- 2) i + (2 y- 2) j + (2 z- 2) k and ∇ g = i + 2 j + 3 k . Setting ∇ f = λ ∇ g gives the equations 2 x- 2 = λ , 2 y- 2 = 2 λ , and 2 z- 2 = 3 λ . Therefore x = λ +2 2 = y +1 2 and z = 3 λ +2 2 = 3 y- 1 2 ....
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HW06 - HW 6 Solutions Math 1920 Spring 2009 March 3 2009...

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