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I am trying to understand Lagrangian multipliers and using an example problem I found online.

Problem Set Up:

Consider a consumer with utility function $u(x,y) = x^{\alpha} y^{1-\alpha}$, where $\alpha \in (0,1)$. Suppose this consumer has wealth $w$ and the prices $p =(p_x,p_y)$. That's all we were given.

Work I Did:

I then defined a budget constraint equation: $w = xp_x + yp_y$. I also then defined an associated Lagrangian for the consumer's maximization problem: $\Lambda(x,y,\lambda) = x^{\alpha} y^{1-\alpha} + \lambda ((xp_x+yp_y)-w)$.

My question:

What does this equation allow me to do? Although I set it up given the formula on Wikipedia's page on Lagrangian multipliers, I really have no idea what the purpose of this equation is. Like I don't understand how the equation as given allows me to determine how to maximize my utility function.

Note: I am familiar with multivariable calculus and Lagrangians ($L = T -V$) in physics, but this method is new to me.

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    $\begingroup$ You might consider asking this at math.stackexchange.com if you do not get a good answer here! Good question. $\endgroup$ – 123 Jan 10 '15 at 19:19
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A constrained optimization function maximizes or minimizes an objective subject to one or more constraints. As I understand it, the Lagrangian multiplier approach transforms a constrained optimization problem (I) into an unconstrained optimization problem (II) where the optimal control values to problem II are also the optimal control values to problem I. Additionally, the the objective functions in problems I and II take the same optimal values. The trick is a clever way of putting the constraints into the objective function directly rather than using them separately.

I agree with your presentation of the consumer's maximization problem: $\Lambda(x,y,\lambda) = x^{\alpha} y^{1-\alpha} + \lambda ((xp_x+yp_y)-w)$.

Now we take the partial derivatives with respect to x an y, set them equal to zero, and then solve for x* and y*.

$0=\partial\Lambda / \partial x = \alpha x^{\alpha -1 } y^{1-\alpha} + \lambda p_x = (\alpha / x ) x^{\alpha } y^{1-\alpha} + \lambda p_x$

$\Rightarrow -\lambda = (\alpha / (x p_x)) x^{\alpha } y^{1-\alpha}$

$0 =\partial\Lambda / \partial y = (1 - \alpha) x^{\alpha} y^{-\alpha} + \lambda p_y = ((1 - \alpha) / y ) x^{\alpha } y^{1-\alpha} + \lambda p_y$

$\Rightarrow -\lambda = ((1- \alpha) / (y p_y)) x^{\alpha } y^{1-\alpha}$

$\Rightarrow (\alpha / (x p_x)) x^{\alpha } y^{1-\alpha} = -\lambda = ((1- \alpha) / (y p_y)) x^{\alpha } y^{1-\alpha}$

$\Rightarrow (\alpha / (x p_x)) = ((1- \alpha) / (y p_y))$

$\Rightarrow ( y p_y ) / (1- \alpha) = (x p_x) / \alpha$ (eqn 1)

Recover the budget constraint equation by taking the partial derivative $\partial\Lambda / \partial \lambda = 0$.

$0 = \partial\Lambda / \partial \lambda = xp_x + yp_y - w \Rightarrow xp_x / w + yp_y /w = 1$ (eqn 2)

We now have two equations and two unknowns (x,y) and can solve for x* and y*.

$\Rightarrow yp_y / w = xp_x/w \cdot (1 / \alpha - 1) = xp_x/w / \alpha - xp_x/w$

$\Rightarrow 1 = yp_y / w + xp_x/w = xp_x/w / \alpha$

$\rightarrow \alpha = xp_x/w$ (result 1)

$\Rightarrow \alpha = xp_x/w = 1 - yp_y /w$

$\rightarrow 1-\alpha =yp_y/w$ (result 2)

Results 1 and 2 form the famous constant expenditure shares result for the Cobb-Douglas utility and production functions. Which can also be explicitly solved for x* and y*: $x^* = \alpha w /p_x$ and $y^* = (1-\alpha) w /p_y$ which are the optimal values for both the Lagrangian and the original problems.

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  • $\begingroup$ In terms of your last sentence, why aren't we solving for $\lambda$ too? I recognize, since $\Lambda(x,y,\lambda)$ is order (aka degree) 1 in $\lambda$, taking the partial derivative $\frac{\partial \Lambda}{\partial \lambda}$ removes $\lambda$ since it's derivative is naturally 1 and thus doesn't end up being a variable. Is this intentional? $\endgroup$ – Stan Shunpike Jan 10 '15 at 19:33
  • $\begingroup$ I expanded the answer and hopefully made this a bit clear. Yes, you do make use of $\partial \Lambda / \partial \lambda$, that's how you recover the budget equation and ultimately solve for optimal values of x and y. But you don't actually choose lambda. You can only choose x and y. $\lambda$ ends up more like a price (a shadow price) than a choice variable. $\endgroup$ – BKay Jan 10 '15 at 19:39
  • $\begingroup$ That cleared it up. Thanks for clarifying. I had worked through an example here: math.stackexchange.com/questions/674/… but somehow actually having numbers confused me. Seeing the variables made more sense. $\endgroup$ – Stan Shunpike Jan 10 '15 at 19:43
  • $\begingroup$ @BKay How do you get $\frac{y p_y}{w}=\frac{x p_x}{w(\alpha - 1)}$? $\endgroup$ – Mathemanic Nov 4 '15 at 3:09
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This is for intuition, not for rigor, and assumes we know which way you would want to deviate from the constraint. Here it is easy; you would want to overspend, so we invoke Lagrange to discipline you to spend $w$ rather than more. Think of the problem in the following steps:

  1. You wish to go out and consume pizza ($x$) and beer ($y$), and ask your parents to borrow a credit card.
  2. Your parents know you, so with the credit card you get the following warning: if you spend more than $w$, we will let our evil neighbour Mr. Lagrange smack your fingers, delivering a pain worth $\lambda$ utility units per dollar you overspend.
  3. Look at the Lagrangian; it is now your utility net of penalty, as a function of pizza ($x$), beer ($y$) and pain ($\lambda\cdot(xp_x+yp_y-w)$). From your point of view, you just maximize this for given $\lambda$ (which means, in particular, that if $\lambda$ is very small, exceeding your budget grossly will be worth a small number of slaps from Mr. Lagrange).
  4. From your parents' point of view, they want to adjust $\lambda$ to the number that makes you voluntarily choose to spend precisely $w$, leaving Mr. Lagrange at bay. (Choosing $\lambda$ higher would lead to you underspending, you could adjust the interpretation accordingly.)
  5. Of course you will then choose precisely the level where you are indifferent between having and not having the bundle of additional consumption & penalty. Hence the shadow price interpretation: $\lambda$ is (more precisely: first-order approximates) how much you would be willing to pay - in the same units as your objective function! to have your budget increased.

As for the suggestion to change sign on the constraint: of course it works mathematically, but I hardly ever use it for instruction purposes; leaving it as it is, $u-\lambda(xp_x+yp_y-w)$ exposes a constraint (which you do not like, it reduces your utility) as equivalent to a tax (which you do not like either, for the same reason). From an economic point of view, you get the idea of the constraint being implemented by a tax, and that is instructive in e.g. modeling Pigouvian taxes internalizing (unwanted negative) externalities.

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Using Lgrange multipliers to optimize a function under constraints is a useful technique, although in the end, it provides additional insights and information. Sticking to the case of equality constraints, the problem

$$\max_{(x,y)} u(x,y) = x^{\alpha} y^{1-\alpha},\;\; \alpha \in (0,1)$$ $$\text {s.t.} \;\;w = p_xx + p_yy$$

can of course be transformed in an unconstrained problem by direct substitution:

$$\max_{y} u(x,y) = \left(\frac {w-yp_y}{p_x}\right)^{\alpha} y^{1-\alpha},\;\; \alpha \in (0,1)$$

But in general, direct substitution can produce cumbersome expressions (especially in dynamic problems), where an algebraic mistake will be easy to make. So the Lagrange method has an advantage here. Moreover, the Lagrange multiplier has a meaningful economic interpretation. In this approach, we define a new variable, say $\lambda$, and we form the "Lagrangean function"

$$\Lambda(x,y,\lambda) = x^{\alpha} y^{1-\alpha} + \lambda (w-p_xx-p_yy)$$

First, note that $\Lambda(x,y,\lambda)$ is equivalent to $u(x,y)$, since the added part to the right is identically zero. Now we maximize the Lagrangean with respect to the two variables and we obtain the first order conditions

$$\frac {\partial u}{\partial x} = \lambda p_x$$

$$\frac {\partial u}{\partial y} = \lambda p_y$$

Equating through $\lambda$, this provides quickly the fundamental relation

$$\frac {\partial u/\partial x} {\partial u/\partial y}= \frac {p_x}{p_y}$$

This optimal relation, together with the budget constraint, provide a two-equation system in two unknowns, and so provide the solution $(x^*, y^*)$ as a function of the exogenous parameters (the utility parameter $\alpha$, the prices $(p_x,p_y)$ and the given wealth $w$).

To determine the value of $\lambda$, multiply each first-order condition throughout by $x$ and $y$ respectively and then sum by sides to get

$$\frac {\partial u}{\partial x}x+\frac {\partial u}{\partial y}y = \lambda (p_xx+p_yy) = \lambda w$$

With utility homogeneous of degree one, as it is the case with Cobb-Douglas functions, we have that

$$\frac {\partial u}{\partial x}x+\frac {\partial u}{\partial y}y = u(x,y)$$

and so at the optimum bundle we have

$$u(x^*,y^*) =\lambda^*w$$

And this is how the Lagrange multiplier acquires an economically meaningful interpretation: its value is the marginal utility of wealth. Now, in the context of ordinal utility, marginal utility is not really meaningful (see also the discussion here). But the above procedure can be applied for example to a cost-minimization problem, where the Lagrange multiplier reflects the increase in total cost by a marginal increase in quantity produced, and so it is the Marginal Cost.

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  • $\begingroup$ This was a great explanation. Question: on Wikipedia's page on Lagrangian multipliers, it states However, not all stationary points yield a solution of the original problem. Thus, the method of Lagrange multipliers yields a necessary condition for optimality in constrained problems. does this mean the term "maximization" is incorrect? Because I thought necessary didn't imply sufficient but the converse did. $\endgroup$ – Stan Shunpike Jan 10 '15 at 19:58
  • $\begingroup$ @StanShunpike Indeed, they are just necessary. They become sufficient when the objective function and the constraints have certain properties. For example, with linear constraints and quasi-concave objective function, they are also sufficient. $\endgroup$ – Alecos Papadopoulos Jan 10 '15 at 20:00
  • $\begingroup$ @AlecosPapadopoulos Another way of writing $u(x^*,y^*)$ is the indirect utility function $v$, correct? Thus, if I'm not mistaken, this is an application of the Envelope Theorem, no? $\endgroup$ – Mathemanic Nov 5 '15 at 3:31
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I'd recommend you to work through this answer paragraph by paragraph, making sure you got each of them in turn, or you will get confused. You may even want to ignore later ones if it's not necessary for your purpose.

The main idea hear is that if the point is conditionall extremum, than it is necessarily a stationary point of the Lagrangian, i.e. such point, that all partial derivatives of the Lagrangian are zero in it. To solve the problem you should identify all stationary points and than find the maximum among them.

However in general this recipe is not entity reliable, as the maximum may not exist. Usually you may verify it's existence with Weierstrass theorem. It requires that fiction is continuous and the set is compact which is the case here. In general it means that you need to check any boundary points of the set in question, points $x= 0$ and points $y = 0$.

In this case your equation is insufficient for solution, as the set you are considering is defined by inequalities rather than equalities. You may point out, that the function is monotonic in $x$ and $y$, so the maximum is on the upper right boundary. Also the utility is 0 if $x = 0$ or $y=0$, while there is feasible points where it is strictly positive, so the maximum can not be attained at either left or lower boundaries. Then this approach is completely justified.

In future you should be aware that problem if such type should be generally solved by applying Kuhn-Tucker Theorem and I recomend you to get acquainted with it after you grasp this material.

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As others have noted, the essence of the Lagrange method is to convert a constrained-extremum problem into a form such that the FOC of the free-extremum problem can be applied. In your setup, you transformed the non-constrained problem ($\max u(x,y)$) to:

$$ \Lambda = x^{\alpha} y^{1-\alpha} + \lambda (w-(xp_x+yp_y)) $$

If you assume that the restriction will be met, that is, that $xp_x+yp_y=w$, then the last term will vanish independently of the value of $\lambda$, so that $\Lambda$ will be identical to $u$. The trick is to treat $\lambda$ as an additional choice variable, thus maximizing $\Lambda(x,y,\lambda)$. Since the first order condition for $\lambda$ is

$$ \frac{\partial Z}{\partial \lambda} = w-(xp_x+yp_y) = 0$$ we can be assured of the satisfaction of the constraint and the disappearance of $\lambda$.

As for the interpretation of $\lambda_i$ (the Lagrange multiplier), in broad economic terms it is the shadow price of the $i$th constraint. In your setup, where there's only a budget constraint, the shadow price is the opportunity cost of the budget constraint, that is, the marginal utility of budget money (income).

Another way to view it is that $\lambda$ measures the sensitivity of $\Lambda$ to changes in the (budget) constraint. In fact in can be proven that

$$ \frac{d\Lambda^*}{dw}=\lambda^* $$

Notice that for this interpretation of $\lambda^*$ to make sense you must always express the constraint as $w-(xp_x+yp_y)$, not as $(xp_x+yp_y)-w$ (like you wrote on your setup).

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