Explanation Given Information: The provided function is f ( x , y ) = 1 − x 2 − y 2 − z 2 subject to z = 2 y . Formula used: The steps for a relative extrema of a function f ( x , y , ... ) subject to the constraint g ( x , y , ... ) = 0 . 1. Construct the Lagrangian function, L ( x , y , ... ) = f ( x , y , .... ) − λ g ( x , y , .... ) , Where λ a new unknown is called a Lagrange multiplier. 2. The relative extrema occur at the critical points of L ( x , y , .... ) . Now, set all the partial derivatives of L ( x , y , ... ) equal to zero and solve the resulting system with the provided constraint equation g ( x , y , ... ) = 0 for the unknowns x , y , ... and λ . Calculation: Start by rewriting the problem with the constraint in the form g ( x , y ) = 0 . Maximize f ( x , y ) = 1 − x 2 − y 2 − z 2 subject to z = 2 y . Here, g ( x , y ) = z − 2 y And, the Lagrangian function is, L ( x , y ) = f ( x , y ) − λ g ( x , y ) = 1 − x 2 − y 2 − z 2 − λ ( z − 2

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ISBN: 9781337291248

Author: Waner, Stefan.

Publisher: Cengage Learning,

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Chapter 8.4, Problem 13E

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To calculate: The maximum value of f(x,y)=1−x2−y2−z2 subject to z=2y.

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Chapter 8.4, Problem 12E

Chapter 8.4, Problem 14E

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Chapter 8 Solutions

Applied Calculus

Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - Prob. 6E Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

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The maximum value of f ( x , y ) = 1 − x 2 − y 2 − z 2 subject to z = 2 y .

Solution Summary: The author calculates the absolute maximum value of f(x,y,mathrm...) subject to z=2y.

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To calculate: The maximum value of f(x,y)=1−x2−y2−z2 subject to z=2y.

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Chapter 8 Solutions