Explanation Given information: The function f ( x , y , z ) = x ( y + z ) lies on the curve of intersection of right circular cylinder x 2 + y 2 = 1 and the hyperbolic cylinder x z = 1 . Calculation: Show the given function as follows: f ( x , y , z ) = x ( y + z ) = x y + x z (1) Find the gradient ( ∇ f ) using Equation (1). ∇ f = ( y + z ) i + x j + x k Show the function f subjected to the constraints as follows: g 1 ( x , y , z ) = x 2 + y 2 − 1 = 0 (2) g 2 ( x , y , z ) = x z − 1 = 0 (3) Find the gradient ( ∇ g 1 ) using Equation (2). ∇ g 1 = 2 x i + 2 y j Find the gradient ( ∇ g 2 ) using Equation (3). ∇ g 2 = z i + x k Find the values of x , y , z , μ , and λ that satisfies the following equation: ∇ f = λ ∇ g 1 + μ ∇ g 2 Substitute ( y + z ) i + x j + x k for ∇ f , 2 x i + 2 y j for ∇ g 1 , and z i + x k for ∇ g 2 . ( y + z ) i + x j + x k = λ ( 2 x i + 2 y j ) + μ ( z i + x k ) y + z = 2 λ x + μ z , x = 2 λ y , and x = μ x Simplify x = μ x as follows: x = μ x x = 0 or μ = 1 Case 1: For x = 0 : If the value of x = 0 , then x z = 1 is not possible. Case 2: If the value of μ = 1 , then y + z = 2 λ x + μ z becomes y + z = 2 λ x + z y = 2 λ x Substitute 2 λ y for x . y = 2 λ ( 2 λ y ) y = 4 λ 2 y y = 0 or 4 λ 2 = 1 For y = 0 : Substitute 0 for y in x 2 + y 2 = 1 . x 2 + 0 = 1 x = ± 1 Substitute 1 for x in x z = 1 . ( 1 ) z = 1 z = 1 Substitute − 1 for x in x z = 1 . ( − 1 ) z = 1 z = − 1 Therefore, the points obtained are ( 1 , 0 , 1 ) and ( − 1 , 0 , − 1 ) . For 4 λ 2 = 1 : λ 2 = 1 4 λ = ± 1 2 For λ = − 1 2 : Substitute − 1 2 for λ in y = 2 λ x . y = 2 ( − 1 2 ) x = − x Substitute − x for y in x 2 + y 2 = 1 . x 2 + x 2 = 1 2 x 2 = 1 x = ± 1 2 Substitute ± 1 2 for x in x z = 1 . ( ± 1 2 ) z = 1 z = ± 2 Therefore, the points obtained are ( 1 2 , − 1 2 , 2 ) and ( − 1 2 , 1 2 , − 2 )

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ISBN: 9780135961308

Author: Hass

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Chapter 14, Problem 87PE

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Calculate the extreme values of the given function.

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Chapter 14, Problem 86PE

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