Lagrangian and Production Maximization with Linear Budgetary Constraints

Real Cobb Douglas Production Function

Mathematical Background:

Multivariable calculus is merely an extension of single-variable calculus.  The multivariate approach allows for the use of more sophisticated models where algebra is insufficient.   Multivariable calculus is used in many fields and disciplines including engineering, physics, social sciences and many others where there is a deterministic system that is the object of mathematical examination.  The many applications of calculus include computer science, medicine, statistics, demography, electromagnetism, and economics.  Calculus has a long history-from the calculations of the volume of pyramids in ancient Egypt to the rate of contagion of the recent Swine Flu.  In particular, any discipline which has change (derivative) or that has an accumulated buildup (integral) can be subject to analysis with calculus.

Economics is one of those disciplines that uses calculus extensively to develop its theory of production.   The basic model in economics is involves two inputs, labor and capital, where their individual properties are homogenous.   These two inputs form the independent varibles in a production function which outputs the total quantity produced of a given product.  There is a curve called an iso-quant curve which describes a fixed quantity to be produced and all the possible combinations of labor and capital that can achieve this output.  There is a cost function that is the sum of the inputs multiplied by the price of inputs to equal the total cost.  There is also an iso-cost curve that is a linear combination of the inputs that results in equal total cost given the price of the inputs.

Mathematical Example:

Definition of Variables:

ScreenHunter_13 Jul. 02 22.53

:  The price of capital (k) is the opportunity cost of capital reflected in its rental rate or depreciation

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:The price of labor (l) is the total compensation paid to worker for an a period of time

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:  Cobb Douglas Production function with decreasing returns to scale

The objective is to maximize the quantity produced subject to the budget constraint given the price of labor and capital.  This problem can be solved using the non-linear method of constrained optimization called the Lagrange method.

1. Set up the Lagrangian function this is the production function minus a proportional quantity of the budget constraint;

ScreenHunter_05 Jul. 02 22.50 (OBJECTIVE FUNCTION)

2. Take the partial derivative of the Lagrangian function with respect to each independent variable including lambda.

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Divide equation (1) and (2) to simplify expression

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Substitute the result from this division into the (3) equation to get the optimum amount of capital

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We can do one final substitution of the optimal capital constraint above into the budget constraint to solve for the optimal amount of labor

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3.  Summarize the solution


L=12, K=16

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and this combination satisfies the labor constraint


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and no other combination of labor and capital, given the budget constraint can achieve a higher level of output.

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5 thoughts on “Lagrangian and Production Maximization with Linear Budgetary Constraints

  1. As a sophomore from University of Ghana,Legon my lecturer test us last semester on question he combined,using cobb douglas and lagragian productiion function to find thhe optimum of capital and labour, though I did lagragian multiplier at the beginning of the semester, he had not taught us cobb Douglas production function by then. I must admit that your presentation of example question is so simple and clear i did not struggle to understand the process. I look forward to learning new topics from your web site.thanks

    • Glad you found it interesting and helpful. There is a button on my blog where you can subscribe to updates. Please click on there to receive any updates. Thanks!

  2. Very nice example, and quite easy to understand. You obviously don’t suffer from the need to impress by showering the page with a lot of math formulas.

    However, I think there are two errors. In partial differential equation 3, the denominator should have a lambda instead of an L.

    Also under item 3. You have L = 12 and K = 16. That should read L = 15 and K = 16.

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