<p><strong>Prices</strong></p>
<p>Prices can be assumed to be positive simply because they typically are positive. There can be some rare examples, such as oil price futures on  <a href="https://www.marketwatch.com/amp/story/oil-prices-went-negative-a-year-ago-heres-what-traders-have-learned-since-11618863839" rel="nofollow noreferrer">April 20, 2020</a>. Nonetheless, such examples are quite rare.</p>
<p>However, explicitly assuming prices are positive is not that common. For example, Varian Microeconomic Analysis 3rd edition does not assumes that for many consumer problems (eg see chapter 9).</p>
<p>Some texts like to also explicitly make prices non-negative, but it is not ubiquitous.</p>
<p><strong>Marginal Utility</strong></p>
<p>This is because typical microeconomic problems assume that ‘more is always better’ for consumer (which implies nonsatiation).</p>
<p>If you assume that more is always better you need to assume <span class="math-container">$U’&gt;0$</span>.</p>
<p>If you choose some utility function for which <span class="math-container">$U’&lt;0$</span> will be violated because this implies that at some point consumer consumes so much that she or he won’t want to consume more.</p>
<hr />
<blockquote>
<p>If the λ exists, don't we just get a solution regardless of all these conditions?</p>
</blockquote>
<p>There are some more conditions that have to be satisfied for optimization problem to have a solution(s) such as that the functions are differentiable etc. However, the assumption above do not have much with existence of solution <em>per se</em> (although depending on exact problem they might be necessary). Rather you would want to impose some restrictions on the problem if they are reasonable to reduce the number of possible solutions.</p>
<p>For example, if prices are very rarely negative it might not be interesting to consider potential solutions where prices are negative.</p>
<p>Likewise, while there are certainly cases where humans more is not always better (eg food consumption), if we look at some composite goods or large variety of goods it’s not unreasonable to assume more is always better. There are probably very few people in the world who would not want to increase consumption of something. Some people might prefer traveling more, spending more time in some nice resort, traveling more comfortably, having better education, more healthcare and so on. Most people will prefer more to less, save some exceptions (e.g. Buddhist monks?).</p>
<p>Consequently, assuming <span class="math-container">$U’&gt;0$</span> is reasonable assumption outside some specialized problems where you might want <strong>not</strong> to assume more is always better. You might not seen problems like that because undergraduate or intermediate books and even many graduate books won’t always cover these more specialized cases, but they certainly exist. So it is possible to find problems where it’s not assumed that marginal utility is always positive, they are just very rare.</p>
<hr />
<p><strong>Conditions for the existence of Lagrangian</strong></p>
<p>Here are the conditions for existence of Lagrangian multiplier (following Sydsaeter et al Further Mathematics for Economic Analysis pp 153-154):</p>
<p>Given a problem:</p>
<p><span class="math-container">$$ \max f(x) \text{s.t.} \begin{cases}
  \displaystyle g_j(\mathbf{x}) =0, j=1,...,r \\
  \displaystyle h_k(\mathbf{x}) \leq 0, j=1,...,s 
\end{cases}$$</span></p>
<p>The Lagrangian multipliers will exist (and thus you could apply Lagrangian) if <span class="math-container">$f,g,h$</span> are all <span class="math-container">$C^1$</span> in some open set <span class="math-container">$A$</span> in <span class="math-container">$\mathbb{R}$</span>, and suppose that <span class="math-container">$\mathbf{x^*}$</span> is a local extreme point in the problem above over A. Then there exist numbers <span class="math-container">$\alpha, \lambda_1,...,\lambda_r, \mu_1,...,\mu_s$</span> that are not all 0, such that:</p>
<ul>
<li><span class="math-container">$\alpha \geq 0$</span></li>
<li><span class="math-container">$\alpha \nabla f(\mathbf{x^*}) = \sum \lambda_j \nabla g_j (\mathbf{x^*}) + \sum \mu_k \nabla h_k (\mathbf{x^*})$</span></li>
<li>for each <span class="math-container">$k=1,...,s$</span> one has <span class="math-container">$\mu_k \geq 0 $</span>, and <span class="math-container">$\mu_k=0$</span> if <span class="math-container">$h_k(\mathbf{x^*})&lt;0$</span></li>
</ul>
<p>So existence of the Lagrangian multiplier does not require objective function to be <em>strictly</em> increasing nor does it require all parameters of the constraint <span class="math-container">$g$</span> and <span class="math-container">$h$</span> to be non-negative.</p>
<p>When it comes to existence of solution Lagrangian multiplier will have solution (see Sydsaeter et al pp 146)  <span class="math-container">$\mathbf{x^*} = (x_1^*,...,x_n^*)$</span> if <span class="math-container">$f, g$</span> and <span class="math-container">$h$</span> are <span class="math-container">$C^1$</span> functions, <span class="math-container">$r&lt;n$</span>, standard Khun-Tucker conditions hold  and constant qualification is satisfied (gradient vectors <span class="math-container">$\nabla g_j (\mathbf{x^*}, 1\leq j \leq m$</span> corresponding to those constraints that are active at  <span class="math-container">$\mathbf{x}^*$</span> are linearly independent, and subject to the above mentioned requirements for existence of multipliers, then there will be some unique numbers <span class="math-container">$\mathbf{\lambda}, \mathbf{\mu}$</span> such that:</p>
<ul>
<li><span class="math-container">$\alpha \nabla f(\mathbf{x^*}) = \sum \lambda_j \nabla g_j (\mathbf{x^*}) + \sum \mu_k \nabla h_k (\mathbf{x^*})$</span></li>
<li><span class="math-container">$\mu_k \geq 0 $</span> and <span class="math-container">$\mu_k = 0$</span> if <span class="math-container">$h_k(\mathbf{x^*})&lt;c_k, k =1,...,R$</span></li>
</ul>
<p>and if the Lagrangain is concave in <span class="math-container">$\mathbf{x}$</span> and there is admissible <span class="math-container">$\mathbf{x^*}$</span> that satisfies conditions above, then <span class="math-container">$\mathbf{x^*}$</span> solves the maximization problem.</p>
<p>Note none of the conditions above require all parameters of <span class="math-container">$h$</span> or <span class="math-container">$g$</span> to be positive or <span class="math-container">$f'_i$</span> to be everywhere strictly non-negative (of course the function needs to be concave in the neighbourhood of some local maximum unless there is corner solution, but there is no reason why it should be non-negative everywhere.</p>

CC BY-SA 4.0
2022-02-21T16:12:23.610Z
Thanks for the answer, but I think the main questions were not addressed, possibly because I couldn't convey that clearly. 

**1.** The assumption of prices being positive is to ensure the maximum exists. (The proof requires the continuity of $U$ as well.) Otherwise, the maximum may or may not exist, and the complete income/wealth may not be consumed. Suppose the prices are a mix of negative and positive values. We can still use the Lagrangian method, right?


**2.** If the preferences are not monotonic, can't we use the Lagrangian method?
not tdm's twin
CC BY-SA 4.0
2022-02-21T16:17:07.830Z
@Gang'sBigBoss 1. you can have problems where all wealth or income is not consumed. Also usually if you specify U’>0, agents will want to consume as much as possible extra assumptions on prices or quantities do not always have to be imposed. As mentioned in the Varian example there are textbooks that won’t make that assumption. You can use lagrangian even if some prices are negative but typically you would just keep prices as unspecified parameters. Note not adding condition P>0 is not the same as assuming prices must be negative
1muflon1
CC BY-SA 4.0
2022-02-21T16:24:41.193Z
2. You can use lagrangian if preferences are not monotonic. You could have some arbitrary utility given by sin(x) (such utility would be nonsensical but you can make any utility you want) and lagrangian could still be used to find maximum utility given constraints you impose. Lagrangian is used even outside economics in physics where you can find all sorts of weird objective functions. The assumptions made on utility are to make the a bit problem more realistic.
1muflon1
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2022-02-22T10:15:52.023Z
Minor comment: Nonsatiation doesn't mean "more is always better". That's monotonicity.
VARulle
CC BY-SA 4.0
2022-02-22T10:17:11.693Z
@VARulle so the Oxford reference is wrong for this one? https://www.oxfordreference.com/view/10.1093/oi/authority.20110803100237926
1muflon1
CC BY-SA 4.0
2022-02-22T12:03:07.583Z
Yes, I think it's wrong. See https://economics.stackexchange.com/questions/25031/whats-the-difference-between-local-non-satiation-and-monotonicity/.
VARulle
CC BY-SA 4.0
2022-02-22T12:39:46.807Z
@VARulle but that answer is about **local** nonsatistion. Local nonsatiation does not require monotonicity but nonsatiation without qualifier local requires monotonicity (local nonsatiation is like a local maximum- it’s just nonsatiation over some range), nonsatiation itself means that more is always better
1muflon1
CC BY-SA 4.0
2022-02-22T13:34:45.820Z
A satiation point is a point such that no other point is preferred. Nonsatiation means that there is no satiation point (assumption III.b in Arrow/Debreu 1954). This is weaker than local nonsatiation.
VARulle
CC BY-SA 4.0
2022-02-22T14:17:55.663Z
@VARulle but if there is no point of satiation then how come that stating ‘more is always better’ would be inaccurate? If there is no point for which consumer would not prefer another point, and by the way also the answer you linked states “ Monotonicity implies local nonsatiation” presumably that can be extended beyond local nonsatiation , what am I missing?
1muflon1
CC BY-SA 4.0
2022-02-22T14:25:45.987Z
In your answer you say, "*typical microeconomic problems assume nonsatiation. That is ‘more is always better’ for consumer.*"
But there are nonsatiated preferences for which more is *not* always better.
VARulle
CC BY-SA 4.0
2022-02-22T15:22:18.357Z
@VARulle I see now, thank you for your comments and great patience I edited my answer
1muflon1
CC BY-SA 4.0
2022-02-21T15:48:50.013Z
2022-02-22T15:21:27.733Z
2022-02-22T15:21:27.733Z
1muflon1
<p>Consider the following utility maximization problem</p>
<p><span class="math-container">$$\max_{x_1,...,x_K} U(x) \\ s.t. \sum_{k=1}^K p_kx_k \leq I \\ x_k \geq 0,$$</span></p>
<p>you can set up a Lagrangian function</p>
<p><span class="math-container">$$\mathcal L(x_1,...,x_K,\lambda,\delta_1,...,\delta_K) = U(x) + \lambda(I - \sum_k p_kx_k) + \sum_k\delta_k x_k$$</span></p>
<p>with the first-order conditions</p>
<p><span class="math-container">$$\frac{\partial \mathcal L}{\partial x_k} = \frac{\partial U}{\partial x_k}-\lambda p_k +\delta_k = 0$$</span></p>
<p>Consider then a candidate for a solution <span class="math-container">$x' = (x'_1,...,x'_K)$</span>. Let us assume that there is some <span class="math-container">$j$</span> such that <span class="math-container">$\partial U(x')/\partial x_j &gt;0$</span>. This states that for some good at the candidate solution the agent is non-satiated. Assuming further that <span class="math-container">$p_j&gt;0$</span> this implies that</p>
<p><span class="math-container">$$\frac{\frac{\partial U(x')}{\partial x_j} + \delta_j}{p_j}=\lambda &gt; 0$$</span></p>
<p>because <span class="math-container">$\delta_j \geq 0$</span> from KKT conditions of solution candidate. This implies that the income constraint must be binding <span class="math-container">$\sum_k p_k x'_k = I$</span>.</p>
<p>Let us now assume further that there is some <span class="math-container">$x'_s$</span> for which <span class="math-container">$p_s'&lt;0$</span> this would then imply that</p>
<p><span class="math-container">$$\frac{\partial U(x')}{\partial x_s}=\lambda p_s -\delta_s,$$</span>
where the LHS is strictly negative because <span class="math-container">$\lambda&gt;0$</span> and <span class="math-container">$\delta_s\geq 0$</span> implying that the marginal utility <span class="math-container">$\frac{\partial U(x')}{\partial x_s}&lt;0$</span>. The intuition here is that negative price introduces a <em>money making machine</em> and unless the agent gets negative marginal utility from <span class="math-container">$x_s$</span> at <span class="math-container">$x'$</span> the agent could relax the income constraint while increasing utility. An example would be a person getting positive marginal utility from working with the price of work being <span class="math-container">$-w$</span> the negative of the wage rate.</p>
<p>If <span class="math-container">$p_s=0$</span> then</p>
<p><span class="math-container">$$\frac{\partial U(x')}{\partial x_s} = -\delta_s$$</span></p>
<p>where <span class="math-container">$\delta_s\geq 0$</span> so either the person would be getting negative marginal utility from <span class="math-container">$x_s$</span> or simply 0. However, if <span class="math-container">$x'_s&gt;0$</span> it is further known that <span class="math-container">$\delta_s=0$</span> implying that <span class="math-container">$\frac{\partial U(x')}{\partial x_s}=0$</span>. The intuition here is simply that if the price is 0 the satisfaction of the income constraint is unaffected by <span class="math-container">$x'_s$</span>. If <span class="math-container">$x'_s&gt;0$</span> the agent can increase or decrease <span class="math-container">$x'_s$</span> still satisfying the constraint <span class="math-container">$x'_s\geq 0$</span>. Hence if the agent has negative marginal utility the agent can decrease <span class="math-container">$x'_s$</span> to get higher utility while still satisfying all constraints. If the agent has positive marginal utility the agent can increase <span class="math-container">$x'_s$</span> to get higher utility again while still satisfying all constraints.</p>

CC BY-SA 4.0
2022-02-21T18:42:57.813Z
+1 because you are highlighting some important points, but note constrained optimization problem can have more than single constraint. For example, the existence of money making machine does not matter if you have additional constraints e.g. some limited storage capacity or output constraints. In public econ literature you will sometimes see problem with per product subsidy that makes price of something effectively negative but with cap on consumption. Thus the above won't necessarily hold once you generalize the problem to have arbitrary number of equality and inequality constraint
1muflon1
CC BY-SA 4.0
2022-02-21T18:31:47.760Z
2022-02-21T18:31:47.760Z
Jesper Hybel
<p>I have seen that the prices and <span class="math-container">$\text{MU}_{i}$</span> are assumed to be positive (or, the preferences monotonic). This is always mentioned when a utility maximization problem is being solved with the Lagrangian method. Is there a reason for that? If the <span class="math-container">$\lambda$</span> exists, don't we just get a solution regardless of all these conditions?</p>

<microeconomics><mathematical-economics><consumer-theory><preferences><optimization>
Conditions to use the Lagrangian method
not tdm's twin
# Conditions to use the Lagrangian method

(Asked by: username_0 on 2022-02-21)

I have seen that the prices and $\text{MU}\_{i}$ are assumed to be positive (or, the preferences monotonic). This is always mentioned when a utility maximization problem is being solved with the Lagrangian method. Is there a reason for that? If the $\lambda$ exists, don't we just get a solution regardless of all these conditions?

**Tags:** microeconomics, mathematical-economics, consumer-theory, preferences, optimization

---

## 2 answers

### Answer by username_1 on 2022-02-21. Score: 2 | Accepted answer

**Prices**

Prices can be assumed to be positive simply because they typically are positive. There can be some rare examples, such as oil price futures on [April 20, 2020](https://www.marketwatch.com/amp/story/oil-prices-went-negative-a-year-ago-heres-what-traders-have-learned-since-11618863839). Nonetheless, such examples are quite rare.

However, explicitly assuming prices are positive is not that common. For example, Varian Microeconomic Analysis 3rd edition does not assumes that for many consumer problems (eg see chapter 9).

Some texts like to also explicitly make prices non-negative, but it is not ubiquitous.

**Marginal Utility**

This is because typical microeconomic problems assume that ‘more is always better’ for consumer (which implies nonsatiation).

If you assume that more is always better you need to assume $U’>0$.

If you choose some utility function for which $U’<0$ will be violated because this implies that at some point consumer consumes so much that she or he won’t want to consume more.

---

> If the λ exists, don't we just get a solution regardless of all these conditions?

There are some more conditions that have to be satisfied for optimization problem to have a solution(s) such as that the functions are differentiable etc. However, the assumption above do not have much with existence of solution *per se* (although depending on exact problem they might be necessary). Rather you would want to impose some restrictions on the problem if they are reasonable to reduce the number of possible solutions.

For example, if prices are very rarely negative it might not be interesting to consider potential solutions where prices are negative.

Likewise, while there are certainly cases where humans more is not always better (eg food consumption), if we look at some composite goods or large variety of goods it’s not unreasonable to assume more is always better. There are probably very few people in the world who would not want to increase consumption of something. Some people might prefer traveling more, spending more time in some nice resort, traveling more comfortably, having better education, more healthcare and so on. Most people will prefer more to less, save some exceptions (e.g. Buddhist monks?).

Consequently, assuming $U’>0$ is reasonable assumption outside some specialized problems where you might want **not** to assume more is always better. You might not seen problems like that because undergraduate or intermediate books and even many graduate books won’t always cover these more specialized cases, but they certainly exist. So it is possible to find problems where it’s not assumed that marginal utility is always positive, they are just very rare.

---

**Conditions for the existence of Lagrangian**

Here are the conditions for existence of Lagrangian multiplier (following Sydsaeter et al Further Mathematics for Economic Analysis pp 153-154):

Given a problem:

$$ \max f(x) \text{s.t.} \begin{cases}
\displaystyle g\_j(\mathbf{x}) =0, j=1,...,r \\
\displaystyle h\_k(\mathbf{x}) \leq 0, j=1,...,s
\end{cases}$$

The Lagrangian multipliers will exist (and thus you could apply Lagrangian) if $f,g,h$ are all $C^1$ in some open set $A$ in $\mathbb{R}$, and suppose that $\mathbf{x^\*}$ is a local extreme point in the problem above over A. Then there exist numbers $\alpha, \lambda\_1,...,\lambda\_r, \mu\_1,...,\mu\_s$ that are not all 0, such that:

* $\alpha \geq 0$
* $\alpha \nabla f(\mathbf{x^\*}) = \sum \lambda\_j \nabla g\_j (\mathbf{x^\*}) + \sum \mu\_k \nabla h\_k (\mathbf{x^\*})$
* for each $k=1,...,s$ one has $\mu\_k \geq 0 $, and $\mu\_k=0$ if $h\_k(\mathbf{x^\*})<0$

So existence of the Lagrangian multiplier does not require objective function to be *strictly* increasing nor does it require all parameters of the constraint $g$ and $h$ to be non-negative.

When it comes to existence of solution Lagrangian multiplier will have solution (see Sydsaeter et al pp 146) $\mathbf{x^\*} = (x\_1^\*,...,x\_n^\*)$ if $f, g$ and $h$ are $C^1$ functions, $r, standard Khun-Tucker conditions hold and constant qualification is satisfied (gradient vectors $\nabla g\_j (\mathbf{x^\*}, 1\leq j \leq m$ corresponding to those constraints that are active at $\mathbf{x}^\*$ are linearly independent, and subject to the above mentioned requirements for existence of multipliers, then there will be some unique numbers $\mathbf{\lambda}, \mathbf{\mu}$ such that:

* $\alpha \nabla f(\mathbf{x^\*}) = \sum \lambda\_j \nabla g\_j (\mathbf{x^\*}) + \sum \mu\_k \nabla h\_k (\mathbf{x^\*})$
* $\mu\_k \geq 0 $ and $\mu\_k = 0$ if $h\_k(\mathbf{x^\*})

and if the Lagrangain is concave in $\mathbf{x}$ and there is admissible $\mathbf{x^\*}$ that satisfies conditions above, then $\mathbf{x^\*}$ solves the maximization problem.

Note none of the conditions above require all parameters of $h$ or $g$ to be positive or $f'\_i$ to be everywhere strictly non-negative (of course the function needs to be concave in the neighbourhood of some local maximum unless there is corner solution, but there is no reason why it should be non-negative everywhere.

> **Comments:**
> **username_0**: Thanks for the answer, but I think the main questions were not addressed, possibly because I couldn't convey that clearly.   **1.** The assumption of prices being positive is to ensure the maximum exists. (The proof requires the continuity of $U$ as well.) Otherwise, the maximum may or may not exist, and the complete income/wealth may not be consumed. Suppose the prices are a mix of negative and positive values. We can still use the Lagrangian method, right?   **2.** If the preferences are not monotonic, can't we use the Lagrangian method?  
> **username_1**: @Gang'sBigBoss 1. you can have problems where all wealth or income is not consumed. Also usually if you specify U’>0, agents will want to consume as much as possible extra assumptions on prices or quantities do not always have to be imposed. As mentioned in the Varian example there are textbooks that won’t make that assumption. You can use lagrangian even if some prices are negative but typically you would just keep prices as unspecified parameters. Note not adding condition P>0 is not the same as assuming prices must be negative  
> **username_1**: 2. You can use lagrangian if preferences are not monotonic. You could have some arbitrary utility given by sin(x) (such utility would be nonsensical but you can make any utility you want) and lagrangian could still be used to find maximum utility given constraints you impose. Lagrangian is used even outside economics in physics where you can find all sorts of weird objective functions. The assumptions made on utility are to make the a bit problem more realistic.  
> **username_2**: Minor comment: Nonsatiation doesn't mean "more is always better". That's monotonicity.  
> **username_1**: @VARulle so the Oxford reference is wrong for this one? https://www.oxfordreference.com/view/10.1093/oi/authority.20110803100237926  
> **username_2**: Yes, I think it's wrong. See https://economics.stackexchange.com/questions/25031/whats-the-difference-between-local-non-satiation-and-monotonicity/.  
> **username_1**: @VARulle but that answer is about **local** nonsatistion. Local nonsatiation does not require monotonicity but nonsatiation without qualifier local requires monotonicity (local nonsatiation is like a local maximum- it’s just nonsatiation over some range), nonsatiation itself means that more is always better  
> **username_2**: A satiation point is a point such that no other point is preferred. Nonsatiation means that there is no satiation point (assumption III.b in Arrow/Debreu 1954). This is weaker than local nonsatiation.  
> **username_1**: @VARulle but if there is no point of satiation then how come that stating ‘more is always better’ would be inaccurate? If there is no point for which consumer would not prefer another point, and by the way also the answer you linked states “ Monotonicity implies local nonsatiation” presumably that can be extended beyond local nonsatiation , what am I missing?  
> **username_2**: In your answer you say, "*typical microeconomic problems assume nonsatiation. That is ‘more is always better’ for consumer.*" But there are nonsatiated preferences for which more is *not* always better.  
> **username_1**: @VARulle I see now, thank you for your comments and great patience I edited my answer  

---

### Answer by username_3 on 2022-02-21. Score: 3

Consider the following utility maximization problem

$$\max\_{x\_1,...,x\_K} U(x) \\ s.t. \sum\_{k=1}^K p\_kx\_k \leq I \\ x\_k \geq 0,$$

you can set up a Lagrangian function

$$\mathcal L(x\_1,...,x\_K,\lambda,\delta\_1,...,\delta\_K) = U(x) + \lambda(I - \sum\_k p\_kx\_k) + \sum\_k\delta\_k x\_k$$

with the first-order conditions

$$\frac{\partial \mathcal L}{\partial x\_k} = \frac{\partial U}{\partial x\_k}-\lambda p\_k +\delta\_k = 0$$

Consider then a candidate for a solution $x' = (x'\_1,...,x'\_K)$. Let us assume that there is some $j$ such that $\partial U(x')/\partial x\_j >0$. This states that for some good at the candidate solution the agent is non-satiated. Assuming further that $p\_j>0$ this implies that

$$\frac{\frac{\partial U(x')}{\partial x\_j} + \delta\_j}{p\_j}=\lambda > 0$$

because $\delta\_j \geq 0$ from KKT conditions of solution candidate. This implies that the income constraint must be binding $\sum\_k p\_k x'\_k = I$.

Let us now assume further that there is some $x'\_s$ for which $p\_s'<0$ this would then imply that

$$\frac{\partial U(x')}{\partial x\_s}=\lambda p\_s -\delta\_s,$$
where the LHS is strictly negative because $\lambda>0$ and $\delta\_s\geq 0$ implying that the marginal utility $\frac{\partial U(x')}{\partial x\_s}<0$. The intuition here is that negative price introduces a *money making machine* and unless the agent gets negative marginal utility from $x\_s$ at $x'$ the agent could relax the income constraint while increasing utility. An example would be a person getting positive marginal utility from working with the price of work being $-w$ the negative of the wage rate.

If $p\_s=0$ then

$$\frac{\partial U(x')}{\partial x\_s} = -\delta\_s$$

where $\delta\_s\geq 0$ so either the person would be getting negative marginal utility from $x\_s$ or simply 0. However, if $x'\_s>0$ it is further known that $\delta\_s=0$ implying that $\frac{\partial U(x')}{\partial x\_s}=0$. The intuition here is simply that if the price is 0 the satisfaction of the income constraint is unaffected by $x'\_s$. If $x'\_s>0$ the agent can increase or decrease $x'\_s$ still satisfying the constraint $x'\_s\geq 0$. Hence if the agent has negative marginal utility the agent can decrease $x'\_s$ to get higher utility while still satisfying all constraints. If the agent has positive marginal utility the agent can increase $x'\_s$ to get higher utility again while still satisfying all constraints.

> **Comments:**
> **username_1**: +1 because you are highlighting some important points, but note constrained optimization problem can have more than single constraint. For example, the existence of money making machine does not matter if you have additional constraints e.g. some limited storage capacity or output constraints. In public econ literature you will sometimes see problem with per product subsidy that makes price of something effectively negative but with cap on consumption. Thus the above won't necessarily hold once you generalize the problem to have arbitrary number of equality and inequality constraint  

---

