Explanation Given: Production function level is to be given P ( x , y ) = 100 x 0.25 y 0.75 where x represents the number of units of labor and y represents the number of units of capital. Total Cost for labor is $ 112 per unit. Cost of capital is $ 60 per unit. Total cost is limited to $ 250 , 000 . Formula Used: Calculation: Objective is to use the method of Lagrange multipliers for determining the maximum production level. Consider the objective function represented by graph z = f ( x , y ) and Constraint equation g ( x , y ) = c where c is constant. Below equations are to be solved by using Lagrange multipliers method for solving constrained optimization problems. f x ( x , y ) = λ g x ( x , y ) f y ( x , y ) = λ g y ( x , y ) g ( x , y ) = c Differentiating the given objective function partially with respect to x , we get P x ( x , y ) = ∂ ∂ x ( 100 x 0.25 y 0.75 ) = 100 ( 0.25 ) x 0.25 − 1 y 0.75 = 25 x − 0.75 y 0.75 = 25 ( y x ) 0.75 ........... ( 1 ) Differentiating the given objective function partially with respect to y , we get P y ( x , y ) = ∂ ∂ y ( 100 x 0.25 y 0.75 ) = 100 ( 0.75 ) x 0.25 y 0.75 − 1 = 75 x 0.25 y − 0.25 = 75 ( x y ) 0.25 ............ ( 2 ) Differentiating the given Constraint function partially with respect to x , we get g x ( x , y ) = ∂ ∂ x ( 112 x + 60 y ) = 112 Differentiating the given constraint function partially with respect to y , we get g y ( x , y ) = ∂ ∂ y ( 112 x + 60 y ) = 60 Substitute the determined partial derivative in the equation (1) of Lagrange multiplier method

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ISBN: 9781337275347

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 13.10, Problem 50E

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Maximum production level using Lagrange multiplier.

Solution Summary: The author calculates the maximum production level using Lagrange multipliers. The objective function is represented by graph z=f(x,y) and Constraint equation

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