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August 21, 2014

August 21, 2014

Posted by **klynn** on Friday, October 5, 2007 at 11:18am.

1.) The mathematics of consumer equilibrium can be demonstrated using the following logic:

A consumer will spend or save his/her entire income, so that:

I = P1X1 + P2X2 + .......... PnXn

Where I = consumer income

X1X2,..............Xn = quantities of n commodities, one of which is saving

P1P2,.............Pn = respective prices of the n commodities (the implied price of saving would be 1)

The consumer wants to maximize his/her total utility, TU, where:

TU = f(X1,X2,..........Xn)

subject to the above income constraint.

a.) Set up the Lagrangian expression that is to be maximized.

I think (?) the answer to this is:

Ltu = TU + lamba(I - P1X1 - P2X2......PnXn)

b.) Show the partial derivative equations whose solution gives optimal values for X1,X2,.........Xn. Note that the partial derivative of TU with respect to X1 is designated as MUx1.

According to my math, these are the partial derivatives:

dLtu/x1 = dTU/dX1 - lambdaP1

This one would be the marginal utility of x1, I think?

dLtu/xn = dTU/dXn - lambdaPn

And, this one shoul dbe the marginal utility of Xn.

c) Show that consumer equilibrium requires the following relationship:

MUx1/P1 = MUx2/P2 = ....... = MUn/Pn

Interpret the economic meaning of the above relationship.

On this one, I'm completely stuck. I've used the Lagrangian Multipler Technique dozens of times, but it's always had actual figures in it. Without actual numbers, I don't even know where to start. :(

- Managerial Economics/Math -
**economyst**, Friday, October 5, 2007 at 1:12pmyour a) and b) look correct.

For c) you are almost there.

You have Ln = dTUn/dXn - LambdaPn

The first term to the right of = is MUn. Also, at a maxima, the whole term is zero. So you have MUn - LambdaPn=0.

Thus, MUn = LambdaPn. This is true for all goods. So, MUx1=LambdaPx1

Thus MUn/MUx1 = LambdaPn/LambdaPx1

The Lambdas cancel. So MUn/MUx1=Pn/Px1

which can be re-written as MUn/Pn = MUx1/Px1

- Managerial Economics/Math -
**klynn**, Tuesday, October 9, 2007 at 2:37pmOk, I think I get it. Thank you so much!

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