Explanation Given: The company can manufacture P ( x , y ) units of a certain product. The labor and the capital are x units and y units respectively. The unit cost of the labor and the capital are p and q ( p > 0 , q > 0 ) respectively. The allocated cost by the management to manufacture the product is C dollars. Here x ∗ and y ∗ are the units of labor and capital and the terms P x ( x ∗ , y ∗ ) and P y ( x ∗ , y ∗ ) are non-zeros. Formula used: Lagrange Multipliers: To obtain the relative extremum of the function f ( x , y ) subject to the constraint g ( x , y ) = 0 (assuming that these extreme values exist), (1) Form an auxiliary function F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) called the Lagrangian function where the variable λ is called the Lagrange multiplier. (2) Solve the system that consists of the equations F x = 0 , F y = 0 and F λ = 0 for all values of x , y , and λ . (3) The solutions found in step (2) are candidates for the extremum of f . Proof: From given p and q are the unit cost of labor and capital in dollars and p ∗ and q ∗ the unit cost of labor and capital with allocated C dollars. Since p is the unit cost of labor, then px be the constrained value of labor. Similarly q is the unit cost of capital, then qy be the constrained value of capital. The constraint equation P ( x ∗ , y ∗ ) in the form g ( x , y ) = p x + q y − C = 0 . The Lagrangian function is as follows. F ( x , y , λ ) = P ( x , y ) + λ g ( x , y ) = P ( x ∗ , y ∗ ) + λ ( p x + q y − C ) The above expression is called the Lagrange function while λ be the Lagrange multipliers. Differentiate F ( x , y , λ ) = P ( x ∗ , y ∗ ) + λ ( p x + q y − C ) with respect to x , d d x ( F ( x , y , λ ) ) = d d x ( P ( x ∗ , y ∗ ) + λ ( p x + q y − C ) ) F x = d d x ( P ( x ∗ , y ∗ ) ) + d d x ( λ ( p x + q y − C ) ) = P x ( x ∗ , y ∗ ) + λ p Differentiate F ( x , y , λ ) = P ( x ∗ , y ∗ ) + λ ( p x + q y − C ) with respect to y , d d y ( F ( x , y , λ ) ) = d d y ( P ( x ∗ , y ∗ ) + λ ( p x + q y − C ) )

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Chapter 8.5, Problem 31E

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To show: The maximized level production Px(x∗,y∗)Py(x∗,y∗)=pq by method of Lagrange multipliers.

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Chapter 8.5, Problem 30E

Chapter 8.5, Problem 33E

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Chapter 8 Solutions

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Compute g(1, 1, 1), g(1,...Ch. 8.1 - Prob. 10E Ch. 8.1 - Prob. 11E Ch. 8.1 - Prob. 12E Ch. 8.1 - In Exercises 11-18, find the domain of the...Ch. 8.1 - Prob. 14E Ch. 8.1 - In Exercises 11-18, find the domain of the...Ch. 8.1 - Prob. 16E Ch. 8.1 - In Exercises 11-18, find the domain of the...Ch. 8.1 - Prob. 18E Ch. 8.1 - Prob. 19E Ch. 8.1 - Prob. 20E Ch. 8.1 - In Exercises 19-24, sketch the level curves of the...Ch. 8.1 - Prob. 22E Ch. 8.1 - In Exercises 1924, sketch the level curves of the...Ch. 8.1 - Prob. 24E Ch. 8.1 - Find an equation of the level curve of f (x, y) =...Ch. 8.1 - Prob. 26E Ch. 8.1 - Prob. 27E Ch. 8.1 - Prob. 28E Ch. 8.1 - Prob. 29E Ch. 8.1 - Prob. 30E Ch. 8.1 - Prob. 31E Ch. 8.1 - Prob. 32E Ch. 8.1 - Prob. 33E Ch. 8.1 - Prob. 34E Ch. 8.1 - Prob. 35E Ch. 8.1 - POISEITLLES LAW Poiseuilles Law states that the...Ch. 8.1 - COST FUNCTION FOR A LOUDSPEAKER SYSTEM Acrosonic...Ch. 8.1 - Prob. 38E Ch. 8.1 - REVENUE FUNCTIONS FOR DESKS Country Workshop...Ch. 8.1 - Prob. 40E Ch. 8.1 - 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The maximized level production P x ( x ∗ , y ∗ ) P y ( x ∗ , y ∗ ) = p q by method of Lagrange multipliers.

Solution Summary: The author explains how Lagrange multipliers can be used to obtain the relative extremum of the function f(x,y,lambda ).

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To show: The maximized level production Px(x∗,y∗)Py(x∗,y∗)=pq by method of Lagrange multipliers.

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Chapter 8 Solutions