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So I have this problem, and what I do is to set up the max problem.( for Lambda Ill use the letter "L" and for what it looks like gamma Ill use letter "g" )

Objective function is Bln(x-t) and the constraint is Px=y

I set up the Lagrangian as follows Bln(x-g)+L(y-Px).

I then write the FOC , so derivatives of the Langrangian in respect to x , y and L.

(1) respect to x => ((b)/(x-g))-L*P = 0

(2) respect to y => L =0

(3) respect to L => y-P*x = 0

So I made it up to here, I thinkI have made some mistakes as equation number 2 doesnt really make sense, I had never done an optimization problem with paramters. Any help please?

Maximilian1988
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  • As $y$ is fixed, differentiating wrt it doesn't make sense, so eqn (2) is not valid. Your other equations also need to be written in terms of $x_i$ as there are $n$ of those variables, not just one $x$. – Macavity Oct 24 '13 at 03:59
  • ok, the way I have set up the LAgrangian is correct? then why I have to write the first equation in terms of xi? there must be only one x that optimize the objective function, it is true that this x is dependent on B and G though , thats why I write X, because thats the endogenous variable, B and G are exogenous..what would be the FOC then? in terms of x and in terms of L only? cheers – Maximilian1988 Oct 25 '13 at 02:53
  • What is $x$? It is not defined in the problem. It is the quantity of what consumed out of the n goods? – Macavity Oct 25 '13 at 03:07
  • yes , X is the amount of good you consume. I understand that there is only one good , defined as X – Maximilian1988 Oct 25 '13 at 04:14
  • The problem (in the picture you posted) however states there are $n$ goods being consumed, and $x_i$ denotes the amount of each. At the optimal state, you will have some value for each $x_i$ to consume. So if thats correct, you cannot replace all of that with one $x$. – Macavity Oct 25 '13 at 04:34
  • Mmm, I think that what it says is that he consumes n goods, of a one same good, like he consumes n potatoes(x)...not that he consumes n quantity of any good. – Maximilian1988 Oct 25 '13 at 08:06
  • Ok, last hint. If there is only one good, there is no optimisation problem, just substitute from the constraint. – Macavity Oct 25 '13 at 08:18

1 Answers1

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Objective function is B*ln(x-t) and the constraint is P*x=y

Either you are ignoring the presence of multiple goods, or are using some vectorized notation. I won't do either, so for me the objective is $$\sum_{i=1}^n \beta_i \ln(x_i-\gamma_i)$$ and the constraint is $$\sum_{i=1}^n p_i x_i = y$$ The Lagrangian is $$L=\sum_{i=1}^n \beta_i \ln(x_i-\gamma_i) + \lambda \left(\sum_{i=1}^n p_i x_i - y \right)$$ By the way, what are we optimizing over? Over amounts $x_i$, I think. The quantity $y$ is not consumer's to choose: of course, if s/he could choose the income to have, the optimization problem becomes pointless: just have yourself a lot of money and buy all you want. This is why taking the derivative with respect to $y$ leads to an absurd equation.

The relevant partial derivatives are those with respect to variables: $x_i$ and the Lagrange multiplier $\lambda$. $$\frac{\partial L}{\partial x_i} = \frac{\beta_i}{x_i-\gamma_i} + \lambda p_i$$ $$\frac{\partial L}{\partial \lambda} = \sum_{i=1}^n p_i x_i - y$$ These should be equated to $0$, etc.

user103402
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