<p>I'm purposefully handwavy here; getting things formally right can be <a href="https://webspace.science.uu.nl/%7Ebalde101/giffenvol.pdf" rel="nofollow noreferrer">very subtle</a>.</p>
<p>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.</p>

CC BY-SA 4.0
2023-05-19T19:53:02.990Z
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!
CormJack
CC BY-SA 4.0
2023-05-19T12:57:09.977Z
2023-05-19T12:57:09.977Z
Michael Greinecker
<p>In general the Lagrange multipliers from a constrained optimization problem takes on the interpretation of the value of adjusting the &quot;size&quot; of the constraint in question.</p>
<p>There can be many interpretations in different contexts, but usually the clearest one (though not clear to beginners) is the &quot;shadow price&quot; of changing the constraint.</p>
<p>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.)</p>
<p>I hope this clarifies things a bit or at the very least does not confuse you further.</p>

CC BY-SA 4.0
2023-05-06T13:47:32.583Z
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?
CormJack
CC BY-SA 4.0
2023-05-04T17:45:19.137Z
2023-05-04T21:44:12.350Z
2023-05-04T21:44:12.350Z
EconJohn
<p>Consider the following constraint optimization with an equality constraint:
<span class="math-container">$$\begin{align}
\max_{(x,y)\in\mathbb{R}^2} \quad &amp; f(x,y) \\
\textrm{s.t.} \quad &amp; g(x,y)=c
\end{align}$$</span></p>
<p>The lagrangian function for the above is: <span class="math-container">$$\mathcal{L}(x,y,\lambda)=f(x,y)-\lambda[g(x,y)-c]$$</span></p>
<p>Let us assume that there exists a solution to the above problem for each <span class="math-container">$c\in\mathbb{R}$</span> and let us denote the solution by <span class="math-container">$(x^*(c),y^*(c),\lambda^*(c))$</span> and further suppose that <span class="math-container">$x^*, y^*, \lambda^*$</span> are differentiable functions.</p>
<p>Since, <span class="math-container">$(x^*,y^*,\lambda^*)$</span> is a solution it must satisfy the following conditions:
<span class="math-container">$$\begin{eqnarray}
&amp; f_1(x^*,y^*)-\lambda^*g_1(x^*,y^*)=0 \tag{1} \\
&amp; f_2(x^*,y^*)-\lambda^*g_2(x^*,y^*)=0 \tag{2}\\
&amp; g(x^*,y^*)-c= 0 \tag{3}
\end{eqnarray}$$</span></p>
<p>where, <span class="math-container">$f_1=\frac{\partial f}{\partial x}$</span> , <span class="math-container">$f_2=\frac{\partial f}{\partial y}$</span> likewise for <span class="math-container">$g_1, g_2$</span></p>
<p>let  <span class="math-container">$h(c)=f(x^*(c),y^*(c))$</span> be the optimal value of the objective function as a function of <span class="math-container">$c$</span></p>
<p>Suppose we are interested in seeing how does the optimum value of the above problem changes in response to a change in <span class="math-container">$c$</span>.</p>
<p>differentiating <span class="math-container">$h(c)$</span> w.r.t. <span class="math-container">$c$</span>: <span class="math-container">$$\begin{eqnarray}
\frac{dh}{dc} &amp; =f_{1}(x^*,y^*)\cdot[x^*(c)]'+ f_{2}(x^*,y^*) \cdot[y^*(c)]' &amp; \quad \text{where }[x^*(c)]'=\frac{dx^*}{dc} , [y^*(c)]'=\frac{dy^*}{dc} \\
&amp;=\lambda^*g_1(x^*,y^*)[x^*(c)]'+ \lambda^*g_2(x^*,y^*)[y^*(c)]' &amp; \quad \text{ from  } (1) \; \&amp; \; (2) \\
&amp;=\lambda^*\{g_1(x^*,y^*)[x^*(c)]'+g_2(x^*,y^*)[y^*(c)]'\}
\end{eqnarray}$$</span></p>
<p>from <span class="math-container">$(3): \quad g(x^*,y^*)=c$</span></p>
<p>differentiating w.r.t. <span class="math-container">$c: \quad g_1(x^*,y^*)[x^*(c)]'+g_2(x^*,y^*)[y^*(c)]'=1$</span></p>
<p>substituting the above in <span class="math-container">$h'(c)$</span>, we get:
<span class="math-container">$$h'(c)=\lambda^*$$</span></p>
<p>We have demonstrated that changing the value of the constraint <span class="math-container">$(c)$</span> leads to a change in the optimal value of the objective function <span class="math-container">$(h(c))$</span> by the value of the Lagrange multiplier<span class="math-container">$(\lambda)$</span>.</p>
<p>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.</p>

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2023-05-20T13:45:31.340Z
Question: why have you used the same function g to denote both the optimal value function as well as the constraint?
Kwame Brown
CC BY-SA 4.0
2023-05-20T16:42:07.200Z
@KwameBrown My bad, slipped my mind and used the same notation. Editing it for clarity, thanks for pointing it out
mynameparv
CC BY-SA 4.0
2023-05-19T12:26:56.240Z
2023-05-20T16:44:47.333Z
2023-05-20T16:44:47.333Z
mynameparv
<p>The Lagrange multiplier is both of the above, both statements are equivalent.</p>
<p>In the classic consumer problem</p>
<p><span class="math-container">$\max U(x,y)$</span></p>
<p>s.t. <span class="math-container">$p_x x + p_y y = I$</span></p>
<p>whose Lagrangian is</p>
<p><span class="math-container">$\mathcal{L}(x,y,\lambda) = U(x,y) + \lambda (I - p_x x - p_y y)$</span>,</p>
<p><span class="math-container">$\lambda$</span> is the shadow price of the income/budget constraint.</p>
<p>The marginal utility/benefit of increasing income is <span class="math-container">$\lambda$</span> utils, i.e, increasing income by <span class="math-container">$\\\$1$</span> yields the individual approximately <span class="math-container">$\lambda$</span> utils if they still spend it optimally.</p>
<p>It can also be viewed as a cost (in terms of utility) of not having more income.</p>

CC BY-SA 4.0
2023-05-06T13:49:46.590Z
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
CormJack
CC BY-SA 4.0
2023-05-06T14:42:59.287Z
Yes, @CormJack you got it!
Nicolas Torres
CC BY-SA 4.0
2023-05-03T12:46:37.430Z
2023-05-03T16:05:53.660Z
2023-05-03T16:05:53.660Z
Nicolas Torres
<p>Is the Lagrange multiplier:</p>
<ol>
<li>The marginal cost of the constraint?</li>
<li>The marginal benefit of relaxing the constraint?</li>
<li>Through duality, both interpretations imply the other?</li>
</ol>
<p>If anyone were so kind, I have another question on Lagrange Multipliers and Non-Negativity constraints <a href="https://economics.stackexchange.com/questions/55274/non-negativity-constraints-kkt">here</a>. Thanks!</p>

<mathematical-economics><optimization><lagrangian>
Lagrange Multiplier Dual Meaning?
CormJack
# Lagrange Multiplier Dual Meaning?

(Asked by: username_0 on 2023-05-03)

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](https://economics.stackexchange.com/questions/55274/non-negativity-constraints-kkt). Thanks!

**Tags:** mathematical-economics, optimization, lagrangian


---

## 4 answers

### Answer by username_5 on 2023-05-03. Score: 6 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks @username_5. 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  
> **username_5**: Yes, @username_0 you got it!  

---

### Answer by username_1 on 2023-05-19. Score: 3

I'm purposefully handwavy here; getting things formally right can be [very subtle](https://webspace.science.uu.nl/%7Ebalde101/giffenvol.pdf).

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.

> **Comments:**
> **username_0**: I appreciate you bringing the intuition to this username_1 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!  

---

### Answer by username_2 on 2023-05-04. Score: 3

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.

> **Comments:**
> **username_0**: Thanks for the answer @username_2, 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?  

---

### Answer by username_3 on 2023-05-19. Score: 3

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.

> **Comments:**
> **username_4**: Question: why have you used the same function g to denote both the optimal value function as well as the constraint?  
> **username_3**: @username_4 My bad, slipped my mind and used the same notation. Editing it for clarity, thanks for pointing it out  

---

