Explanation Lagrange Multiplier is used for solving the optimization of the function with some constraints. Given: Given functions is f ( x , y , z ) = x + 2 y + 3 z and constraints are x + y + z = 1 and x 2 + y 2 + z 2 = 1 Formula Used: According to Lagrange Multiplier, let us suppose that we have three differentiable functions f ( x , y ) , g ( x , y ) and h ( x , y ) , if f ( x , y ) has a local maximum or a local minimum on the constraint curve g ( x , y ) = 0 and h ( x , y ) = 0 at a point P ( a , b ) and if ∇ g p ≠ 0 and ∇ h P ≠ 0 , then there exists a scalar quantity λ such that ∇ f P = λ ∇ g P + μ ∇ h P Calculation: Given the constraint curves are x + y + z = 1 and x 2 + y 2 + z 2 = 1 Put g ( x ) = 0 and h ( x ) = 0 Thus, we have g ( x ) = x + y + z - 1 = 0 and h ( x ) = x 2 + y 2 + z 2 - 1 = 0 Gradient of f ( x , y , z ) = x + 2 y + 3 z is given as: ∇ f = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) ∂ f ∂ x ( x + 2 y + 3 z ) = 1 + 2 ( 0 ) + 3 ( 0 ) = 1 ∂ f ∂ y ( x + 2 y + 3 z ) = (0) + 2 ( 1 ) + 3 ( 0 ) = 2 ∂ f ∂ z ( x + 2 y + 3 z ) = (0) + 2 ( 0 ) + 3 ( 1 ) = 3 Thus, the gradient of the function is ∇ f = 〈 1 , 2 , 3 〉 And, Gradient of g ( x ) = x + y + z - 1 is given as: ∇ g = ( ∂ g ∂ x , ∂ g ∂ y , ∂ g ...

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 14, Problem 59CRE

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The extreme values of the function $f (x, y, z) = x + 2 y + 3 z$ subject to the constraints $x + y + z = 1$ and $x^{2} + y^{2} + z^{2} = 1$

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Chapter 14, Problem 58CRE

Chapter 14, Problem 60CRE

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Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter | (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) ☐ 1. For all n > 1, seriesΣ In(n) In(n) converges. 2, 1, arctan(n) the series arctan(n) n³ ☐ 4. For all n > 1, 123 converges. 1 n ln(n) series In(n) diverges. 2n . and the seriesΣconverges, so by the Comparison Test, 2, 3, and the series converges, so by the Comparison Test, the series-3 1 converges. ☐ 6. For all n > 2, In(n) >, and the series Σ converges, so by the Comparison Test, the seriesΣ In(n) converges.

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Chapter 14 Solutions

Calculus: Early Transcendentals

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 extreme values of the function f ( x , y , z ) = x + 2 y + 3 z subject to the constraints x + y + z = 1 and x 2 + y 2 + z 2 = 1

The extreme values of the function f(x, y,z) = x+2y+3z subject to the constraints x + y + z= 1 and x2 + y2 + z2= 1

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Chapter 14 Solutions

The extreme values of the function $f (x, y, z) = x + 2 y + 3 z$ subject to the constraints $x + y + z = 1$ and $x^{2} + y^{2} + z^{2} = 1$