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Use the method of Lagrange multipliers to optimize $f$ as indicated, subject to the given constraint(s).Production is modeled by the Cobb Douglas production function $f(x, y)=200 x^{2} y^{\frac{1}{5}},$ where $x$ represents the number of units of labor, and $y$ the number of units of capital. If each unit of labor costs $\$ 500$ and each unit of capital $\$ 200,$ and the amount allocated to labor and capital is $\$ 300,000$ use Lagrange multipliers to determine the number of units of labor and capital which maximizes the level of production.

$(400,500)$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 4

The Method of Lagrange Multipliers

Partial Derivatives

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Use the method of Lagrange…

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Use Lagrange multipliers t…

And this problem that give us the so called Cobb Douglas production function. I don't really know the history of that. I'm not economist. Um so but they gave it to us as 200 times X. The two thirds times wider. The one third where X. Is the number of units of labor and why is the number of units of capital? So I don't know where this came from but I guess they basically looked at a lot of mhm. I looked at a lot of businesses and decided that that probably fit that the data. So We're told that Labour costs 500 dollars for each unit and capital costs $200 for each unit. And we have a total amount of money Of $300,000. So our constraint function is C equals 500 X plus 200 Y minus 30 100,000. So we can form our augmented function here with this and this and our lagrange multiplier take partial with respect to X. Y. And X. Y. And lambda. And I think I did not update what, you know I did. Okay um in the in the on the website the problem they have these exponents wrong. So I had originally solved it with the wrong exponents and I thought maybe I hadn't updated the values here. So um we just get three equations, three unknowns. We can solve this for um what did I do? Um Yeah I forgot to put in when I corrected this, forgot to put in my the minus 500 lambda equals zero And plus 200 And that equals zero. All right. So those are three equations. You know, we can basically solve this lambda in terms of X and Y and get rid of it in here. And then we have two equations for X and Y. And we can solve those And we've come out with is that x equals 400 and what I equals 500. So you want 400 units of labor and 500 units of capital. And then we get a value for λ here, that is kind of irrelevant. Um So what we wind up with this is our solution here and we can we can kind of do this the other way of taking solving for this in terms of so this is F. Bar and I saw I saw that in terms for all this is X. I saw this for why? In terms of acts plugged it into here and then plotted it. And we get this interesting looking function here and we can see our maximum occurs right around there. So we have zero out here. But we have where we obviously we have we don't have zero labor And if we go out here just 600 units of labor, then we have no capital left. So that's where we get zero these two zeros. And in between there, you know, we We have this maximum where x equals 400. And then that would tell us why equals 500

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