Explanation Method used: In order to find the maximum and minimum values of a function f ( x , y , z ) that is subjected to the constraint g ( x , y , z ) = k such that ∇ g ≠ 0 . a) Find the values x, y , z and λ when ∇ f ( x , y , z ) = λ ∇ g ( x , y , z ) and g ( x , y , z ) = k . b) Compute the values at the points from step a and find the maximum and minimum values of f . Calculation: It is given that the function is f ( x , y , z ) = 3 x − y − 3 z and the constraints are x + y − z = 0 and x 2 + 2 z 2 = 1 . Here, g ( x , y , z ) = x + y − z and h ( x , y , z ) = x 2 + 2 z 2 . Find f x as f x ( x , y , z ) = 3 , f y as f y ( x , y , z ) = − 1 , f z as f z ( x , y , z ) = − 3 , g x as g x ( x , y , z ) = 1 , g y as g y ( x , y , z ) = 1 , g z as g z ( x , y , z ) = − 1 , h x as h x ( x , y , z ) = 1 , h y as h y ( x , y , z ) = 0 , and h z as h z ( x , y , z ) = 4 z . Thus, the equations are obtained as f x ( x , y , z ) = λ g x ( x , y , z ) + μ h x ( x , y , z ) 3 = λ ( 1 ) + μ ( 2 x ) 3 = λ + 2 μ x f y ( x , y , z ) = λ g y ( x , y , z ) + μ h y ( x , y , z ) − 1 = λ ( 1 ) + μ ( 0 ) λ = − 1 Then from 3 = λ + 2 μ x , the equation is obtained as 3 = − 1 + 2 μ x 4 = 2 μ x μ x = 2 μ = 2 x . f z ( x , y , z ) = λ g z ( x , y , z ) + μ h z ( x , y , z ) − 3 = ( − 1 ) ( − 1 ) + μ ( 4 z ) − 3 = 1 + 4 μ z − 4 = 4 μ z That is, μ z = − 1 μ = − 1 z Equate the values of μ : 2 x = − 1 z z = − x 2 Substitute the value of z in the equation x 2 + 2 z 2 = 1 and obtain the value of x as follows. x 2 + 2 ( − x 2 ) 2 = 1 x 2 + x 2 2 = 1 3 2 x 2 = 1 x = ± 2 3 Then, the value of z becomes z = − 1 2 ( ± 2 3 ) = ∓ 1 2 2 3

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

2nd Edition

ISBN: 9781285101552

Author: James Stewart

Publisher: Cenage

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Chapter 11.8, Problem 14E

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To find: The extreme values of the function f that is subjected to the given constraints.

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Chapter 11 Solutions

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

Ch. 11.1 - Let g(x, y) = cos(x + 2y). (a) Evaluate g(2, 1)....Ch. 11.1 - Let F(x,y)=1+4y2. (a) Evaluate F(3,1). (b) Find...Ch. 11.1 - Let f(x,y,z)=x+y+z+ln(4x2y2z2). (a) Evaluate f(1,...Ch. 11.1 - Let g(x,y,z)=x3y2z10xyz. (a) Evaluate g(1, 2, 3)....Ch. 11.1 - Find and sketch the domain of the function....Ch. 11.1 - Find and sketch the domain of the function....Ch. 11.1 - Find and sketch the domain of the function. 15....Ch. 11.1 - Find and sketch the domain of the function....Ch. 11.1 - Find and sketch the domain of the function. 19....Ch. 11.1 - Find and sketch the domain of the function. f(x,...

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The extreme values of the function f that is subjected to the given constraints.

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Chapter 11 Solutions