Explanation Lagrange Multiplier for solving the optimization of the function with some constraints. Given: Given function is f ( x , y ) = x 2 y and constraint is 4 x 2 + 9 y 2 = 36 Formula Used: According to Lagrange Multiplier, let us suppose that we have two differentiable functions f ( x , y ) and g ( x , y ) , if f ( x , y ) has a local maximum or a local minimum on the constraint curve g ( x , y ) = 0 at a point P ( a , b ) and if ∇ g p ≠ 0 , then there exists a scalar quantity λ such that ∇ f P = λ ∇ g P Calculation: Given the constraint curve 4 x 2 + 9 y 2 = 36 Put g ( x ) = 0 Thus, we have 4 x 2 + 9 y 2 - 36 = 0 Gradient of f ( x , y ) = x 2 y is given as: ∇ f = ( ∂ f ∂ x , ∂ f ∂ y ) ∂ f ∂ x ( x 2 y ) = 2 y x ∂ f ∂ y ( x 2 y ) = x 2 Thus, the gradient of the function is ∇ f = 〈 2 y x , x 2 〉 And, Gradient of g ( x , y ) = 4 x 2 + 9 y 2 - 36 is given as: ∇ g = ( ∂ g ∂ x , ∂ g ∂ y ) ∂ g ∂ x ( 4 x 2 + 9 y 2 - 36 ) = 4 ( 2 x ) + 9 ( 0 ) − 36 ( 0 ) = 8 x ∂ g ∂ y ( 4 x 2 + 9 y 2 - 36 ) = 4 ( 0 ) + 9 ( 2 y ) − 36 ( 0 ) = 18 y Thus, the gradient of the function is ∇ g = 〈 8 x , 18 y 〉 According to Lagrange Theorem, we have ∇ f P = λ ∇ g P Substituting the values, we get 〈 2 y x , x 2 〉 = λ 〈 8 x , 18 y 〉 Thus, we have 2 x y = λ 8 x And x 2 = λ 18 y y = 4 λ And λ = x 2 18 y λ = y 4 Solving the equations x 2 18 y = y 4 4 x 2 = 18 y 2 2 x 2 = 9 y 2 y 2 = 2 x 2 9 y = ± x 2 3 Substituting the values of y 2 in 4 x 2 + 9 y 2 - 36 = 0 , we get 4 x 2 + 9 ...

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ISBN: 9781319379421

Author: Rogawski

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Chapter 14, Problem 57CRE

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The maximum and minimum values of the function $f (x, y) = x^{2} y$ on the ellipse $4 x^{2} + 9 y^{2} = 36$

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Chapter 14, Problem 56CRE

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Chapter 14 Solutions

CALCULUS 4E (HC) W/ ACHIEVE ACCESS

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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The maximum and minimum values of the function f ( x , y ) = x 2 y on the ellipse 4 x 2 + 9 y 2 = 36

The maximum and minimum values of the function f(x, y) = x2y on the ellipse 4x2 + 9y2 = 36

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Chapter 14 Solutions

The maximum and minimum values of the function $f (x, y) = x^{2} y$ on the ellipse $4 x^{2} + 9 y^{2} = 36$