Explanation Given: The given function is f ( x , y ) = x 2 + y 2 . The constraint is g ( x , y ) = x 4 + y 4 − 2 . Formula used: Lagrange function is, L ( x , y ) = f ( x , y ) − λ g ( x , y ) Calculation: Consider the Lagrange function for the given function with multiplier λ . L ( x , y ) = x 2 + y 2 − λ ( x 4 + y 4 − 2 ) L ( x , y ) = x 2 + y 2 − x 4 λ − y 4 λ + 2 λ The 1 st order partial derivatives are as below, L x = 2 x − 4 x 3 λ L y = 2 y − 4 y 3 λ The system equations are, L x = 0 ; L y = 0 ; g ( x , y ) = 0 The above system of equations gives the following results. 2 x − 4 x 3 λ = 0 ⇒ x = 0 , 2 x = 4 x 3 λ ⇒ x = 0 , x 2 = 1 2 λ ⇒ x = ± 1 2 λ , x = 0 2 y − 4 y 3 λ = 0 ⇒ y = 0 , 2 y = 4 y 3 λ ⇒ y = 0 , y 2 = 1 2 λ ⇒ y = ± 1 2 λ , y = 0 Substitute the values of x and y in equation x 4 + y 4 − 2 = 0

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Author: DA

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Chapter 8.6, Problem 10P

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The maximum and minimum value of the function f(x,y)=x2+y2.

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Chapter 8.6, Problem 9P

Chapter 8.6, Problem 11P

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Chapter 8 Solutions

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The maximum and minimum value of the function f ( x , y ) = x 2 + y 2 .

Solution Summary: The author calculates the Lagrange function for the given function with multiplier lambda .

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To find:

The maximum and minimum value of the function f(x,y)=x2+y2.

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Chapter 8 Solutions

The maximum and minimum value of the function f ( x , y ) = x 2 + y 2 .

Solution Summary: The author calculates the Lagrange function for the given function with multiplier lambda .

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To find: The maximum and minimum value of the function f(x,y)=x2+y2.

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Chapter 8 Solutions

To find:

The maximum and minimum value of the function f(x,y)=x2+y2.