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Is the Lagrange multiplier:

  1. The marginal cost of the constraint?
  2. The marginal benefit of relaxing the constraint?
  3. Through duality, both interpretations imply the other?

If anyone were so kind, I have another question on Lagrange Multipliers and Non-Negativity constraints here. Thanks!

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4 Answers 4

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The Lagrange multiplier is both of the above, both statements are equivalent.

In the classic consumer problem

$\max U(x,y)$

s.t. $p_x x + p_y y = I$

whose Lagrangian is

$\mathcal{L}(x,y,\lambda) = U(x,y) + \lambda (I - p_x x - p_y y)$,

$\lambda$ is the shadow price of the income/budget constraint.

The marginal utility/benefit of increasing income is $\lambda$ utils, i.e, increasing income by $\\\$1$ yields the individual approximately $\lambda$ utils if they still spend it optimally.

It can also be viewed as a cost (in terms of utility) of not having more income.

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  • $\begingroup$ Thanks @Nicolas. To be clear if i had a maximisation problem i.e. Max $U$ s.t. budget $m$, both interpretations would be correct here for the maximisation case? I.e. the MC of having the constraint = MB of relaxing the constraint = λ $\endgroup$
    – CormJack
    Commented May 6, 2023 at 13:49
  • $\begingroup$ Yes, @CormJack you got it! $\endgroup$ Commented May 6, 2023 at 14:42
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In general the Lagrange multipliers from a constrained optimization problem takes on the interpretation of the value of adjusting the "size" of the constraint in question.

There can be many interpretations in different contexts, but usually the clearest one (though not clear to beginners) is the "shadow price" of changing the constraint.

To illustrate with your standard utility maximization problem, the multiplier takes on the interpretation of the marginal utility from money but that in and of itself is a shadow price on income (i.e. if there was a market for income that can be purchased with utility it would be at the value of the multiplier.)

I hope this clarifies things a bit or at the very least does not confuse you further.

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  • $\begingroup$ Thanks for the answer @EconJohn, yes i understand your point about "marginal utility from money," i.e. by saying that it's the marginal benefit from increasing the constraint e.g. having more income. Are you also saying that if we had a minimisation problem, where are constraint was a certain level of utility. Then here λ is actually 1/λ from the the maximisation case. But if this Minimisation problem of income in terms of utility was a maximisation problem, i.e. how much money utility we spend to get a desired level of income. It would be the same λ. Is that kind of correct? $\endgroup$
    – CormJack
    Commented May 6, 2023 at 13:47
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I'm purposefully handwavy here; getting things formally right can be very subtle.

Suppose you could ignore the constraint. Then you do your usual unconstrained maximization problem, and the marginal benefit of changing any variable under control is zero. But often, you cannot ignore the constraint. The idea is to modify the problem so that the problem becomes unconstrained but with a maximum that satisfies the constraint. To do this, you introduce a cost for violating the constraint. To do this locally with everything nice and smooth, you can just multiply the deviation from the constraint with a number, the Lagrange multiplier. How large should this cost be? Well, if the constraint should hold at an optimum, the marginal cost should exactly equal the marginal benefit of violating the constraint. And this is why the two interpretations are really the same.

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  • $\begingroup$ I appreciate you bringing the intuition to this Michael thank you! I find somethings make intuitive sense, and then i almost think about them too much, and they stop making sense. I've recently had this with homothetic functions! $\endgroup$
    – CormJack
    Commented May 19, 2023 at 19:53
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Consider the following constraint optimization with an equality constraint: $$\begin{align} \max_{(x,y)\in\mathbb{R}^2} \quad & f(x,y) \\ \textrm{s.t.} \quad & g(x,y)=c \end{align}$$

The lagrangian function for the above is: $$\mathcal{L}(x,y,\lambda)=f(x,y)-\lambda[g(x,y)-c]$$

Let us assume that there exists a solution to the above problem for each $c\in\mathbb{R}$ and let us denote the solution by $(x^*(c),y^*(c),\lambda^*(c))$ and further suppose that $x^*, y^*, \lambda^*$ are differentiable functions.

Since, $(x^*,y^*,\lambda^*)$ is a solution it must satisfy the following conditions: $$\begin{eqnarray} & f_1(x^*,y^*)-\lambda^*g_1(x^*,y^*)=0 \tag{1} \\ & f_2(x^*,y^*)-\lambda^*g_2(x^*,y^*)=0 \tag{2}\\ & g(x^*,y^*)-c= 0 \tag{3} \end{eqnarray}$$

where, $f_1=\frac{\partial f}{\partial x}$ , $f_2=\frac{\partial f}{\partial y}$ likewise for $g_1, g_2$

let $h(c)=f(x^*(c),y^*(c))$ be the optimal value of the objective function as a function of $c$

Suppose we are interested in seeing how does the optimum value of the above problem changes in response to a change in $c$.

differentiating $h(c)$ w.r.t. $c$: $$\begin{eqnarray} \frac{dh}{dc} & =f_{1}(x^*,y^*)\cdot[x^*(c)]'+ f_{2}(x^*,y^*) \cdot[y^*(c)]' & \quad \text{where }[x^*(c)]'=\frac{dx^*}{dc} , [y^*(c)]'=\frac{dy^*}{dc} \\ &=\lambda^*g_1(x^*,y^*)[x^*(c)]'+ \lambda^*g_2(x^*,y^*)[y^*(c)]' & \quad \text{ from } (1) \; \& \; (2) \\ &=\lambda^*\{g_1(x^*,y^*)[x^*(c)]'+g_2(x^*,y^*)[y^*(c)]'\} \end{eqnarray}$$

from $(3): \quad g(x^*,y^*)=c$

differentiating w.r.t. $c: \quad g_1(x^*,y^*)[x^*(c)]'+g_2(x^*,y^*)[y^*(c)]'=1$

substituting the above in $h'(c)$, we get: $$h'(c)=\lambda^*$$

We have demonstrated that changing the value of the constraint $(c)$ leads to a change in the optimal value of the objective function $(h(c))$ by the value of the Lagrange multiplier$(\lambda)$.

The above result has many useful applications. For example, for some standard UMP if a rise in the price of a single commodity leads to a fall in the utility of the consumer assuming everything else stays constant, then using the above result you can figure out much money income we need to give to the consumer to make him as well off as before the increase in prices.

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    $\begingroup$ Question: why have you used the same function g to denote both the optimal value function as well as the constraint? $\endgroup$ Commented May 20, 2023 at 13:45
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    $\begingroup$ @KwameBrown My bad, slipped my mind and used the same notation. Editing it for clarity, thanks for pointing it out $\endgroup$
    – mynameparv
    Commented May 20, 2023 at 16:42

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