Explanation Lagrange Multipliers: If f ( x , y ) and g ( x , y ) are two differentiable functions and f ( x , y ) has a local minimum or local maximum on the constraint curve g ( x , y ) = 0 at P = ( a , b ) , and if ∇ g p ≠ 0 , then there is a scalar λ such that ∇ f = λ ∇ g Here, λ is called the Lagrange’s Multiplier. Given: The function is f ( x , y ) = ( x 2 + 1 ) y and the constraint is x 2 + y 2 = 5 . Formula Used: The Lagrange equation is: ∇ f = λ ∇ g … ( 1 ) Calculation: Consider the given function: f ( x , y ) = ( x 2 + 1 ) y = x 2 y + y The constraint curve is: g ( x , y ) = x 2 + y 2 − 5 = 0 Here, the function is f ( x , y ) = x 2 y + y Then, f x ( x , y ) = ∂ ∂ x ( x 2 y + y ) = 2 x y f y ( x , y ) = ∂ ∂ y ( x 2 y + y ) = x 2 + 1 Now, we have ∇ f = 〈 2 x y , x 2 + 1 〉 Here, g ( x , y ) = x 2 + y 2 − 5 Then, g x ( x , y ) = ∂ ∂ x ( x 2 + y 2 − 5 ) = 2 x g y ( x , y ) = ∂ ∂ y ( x 2 + y 2 − 5 ) = 2 y Now, we have ∇ g = 〈 2 x , 2 y 〉 Then, the Lagrange equation, from the equation (1) is: ∇ f = λ ∇ g Then, 〈 2 x y , x 2 + 1 〉 = λ 〈 2 x , 2 y 〉 Now, we get 2 x y = 2 λ x … ( 1 ) x 2 + 1 = 2 λ y … ( 2 ) Here, if y = 0 , then x 2 + 1 = 0 which is an absurdity...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 15.8, Problem 3E

To determine

To calculate:

The minimum and maximum value of $f (x, y) = (x^{2} + 1) y$ by applying the Lagrange multipliers, subject to the constraint $x^{2} + y^{2} = 5$ .

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Chapter 15.8, Problem 2E

Chapter 15.8, Problem 4E

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Chapter 15 Solutions

CALCULUS (CLOTH)

Ch. 15.1 - Prob. 1PQ Ch. 15.1 - Prob. 2PQ Ch. 15.1 - Prob. 3PQ Ch. 15.1 - Prob. 4PQ Ch. 15.1 - Prob. 5PQ Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - Prob. 3E Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E

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The minimum and maximum value of f ( x , y ) = ( x 2 + 1 ) y by applying the Lagrange multipliers, subject to the constraint x 2 + y 2 = 5 .

Solution Summary: The Lagrange Multiplier calculates the minimum and maximum value of f (x, y) = x 2 + 1.

To calculate:

The minimum and maximum value of $f (x, y) = (x^{2} + 1) y$ by applying the Lagrange multipliers, subject to the constraint $x^{2} + y^{2} = 5$ .

Want to see the full answer?

Chapter 15 Solutions

The minimum and maximum value of f ( x , y ) = ( x 2 + 1 ) y by applying the Lagrange multipliers, subject to the constraint x 2 + y 2 = 5 .

Solution Summary: The Lagrange Multiplier calculates the minimum and maximum value of f (x, y) = x 2 + 1.

To calculate: The minimum and maximum value of f(x,y)=(x2+1)y by applying the Lagrange multipliers, subject to the constraint x2+y2=5.

Want to see the full answer?

Chapter 15 Solutions

To calculate:

The minimum and maximum value of $f (x, y) = (x^{2} + 1) y$ by applying the Lagrange multipliers, subject to the constraint $x^{2} + y^{2} = 5$ .