<p>Remember what is the <strong>goal of optimization</strong>: It is to find some maximum or minimum of a function. Potentially, to find maximum or minimum such that something must be satisfied (constraints)</p>
<p>First order conditions tell you that the directional slopes (partial derivations) of function are <span class="math-container">$0$</span>. If you think about it, slope being equal to <span class="math-container">$0$</span> indicates local maximum or minimum (the function either stops increasing or decreasing).</p>
<p>Now, usually the result of taking FOC of some problem results in <strong>functions</strong> (or <strong>conditions</strong> that must hold for a choice to be optimum), let's call them <span class="math-container">$f'(\boldsymbol{x}) = 0$</span>. If you plug concrete numbers into <span class="math-container">$\boldsymbol{x}$</span>, for example <span class="math-container">$x = 0$</span>, and with that your <span class="math-container">$f'(\boldsymbol{x}) \neq 0$</span>, you know that the condition resulting from FOC <strong>cannot be satisfied by that concrete</strong> <span class="math-container">$\boldsymbol{x}$</span>!</p>
<p>Only such choice <span class="math-container">$\boldsymbol{x}$</span> for which it holds <span class="math-container">$f'(\boldsymbol{x}) = 0$</span> can be optimal!</p>

CC BY-SA 4.0
2023-02-28T08:07:55.473Z
This is not really true since it ignores the possibility of a boundary maximum. $f$ could be strictly decreasing.
VARulle
CC BY-SA 4.0
2023-02-28T08:28:45.643Z
Could you elaborate further? When doing Lagrangian (or KKT when necessary), even boundary maximum requires that $f'(\boldsymbol{x}) = 0$, where $f$ is the Lagrange function.
Athaeneus
CC BY-SA 4.0
2023-02-28T14:16:03.877Z
The $f$ you describe in your answer is defined on "numbers" $x$ and has a derivative denoted by the prime. This cannot be a Lagrange function.
VARulle
CC BY-SA 4.0
2023-02-28T14:41:44.537Z
It's more of a notation thing (by the way, the $x$ is bold in my text). If you read the text, you can see that I speak of conditions or functions and I just denote them by $f'()$. The $\boldsymbol{x}$  is vector of variables, not single "numbers", that is why it is bold. What concerns the rest, the $f'(\boldsymbol{x})$ is a vector of FOC. Just a reminder that I consider the multiplicator to be variable as well.
Athaeneus
CC BY-SA 4.0
2023-03-01T17:35:23.910Z
O.k., but then the prime in $f'({\bf x}) \ne 0$ has to be interpreted as a gradient and the $0$ as the zero vector. And this while the problem of the OP is obviously 1-dimensional...
VARulle
CC BY-SA 4.0
2023-02-28T06:25:58.463Z
2023-02-28T06:25:58.463Z
Athaeneus
<p>If the question asks you to prove <span class="math-container">$x&gt;0$</span>, then the easiest you can do is to come up with the contradiction. Take <span class="math-container">$x=0$</span> and some <span class="math-container">$x&gt;0$</span>, show that the utility <span class="math-container">$u(x&gt;0) &gt; u(x=0)$</span>. You are done.</p>
<p>Analytically, you could investigate your utility function. If it is increasing monotonically in <span class="math-container">$x$</span>, then <span class="math-container">$x$</span> is going to hit the maximum possible constraint, unless you have a constraint that imposes exogenously <span class="math-container">$x=0$</span>.</p>

CC BY-SA 4.0
2023-02-28T16:13:36.533Z
2023-02-28T16:13:36.533Z
Econfox
<p>It's not quite possible to have &quot;FOC greater than 0&quot;, since the FOC is a <em>condition</em>, not a number. What is really meant here is the derivative of the utility function.</p>
<p>If your utility function is given by <span class="math-container">$u(x)$</span>, then the FOC for a maximum would be <span class="math-container">$u'(x)=0$</span>. Now if you calculate the derivative <span class="math-container">$u'(x)$</span> and then substitute <span class="math-container">$0$</span> for <span class="math-container">$x$</span>, then you get the derivative at <span class="math-container">$0$</span>, written <span class="math-container">$u'(0)$</span>.</p>
<p>If it turns out that <span class="math-container">$u'(0)&gt;0$</span>, then this means that your utility function is <em>locally increasing</em> at <span class="math-container">$0$</span>. Hence, if you increase <span class="math-container">$x$</span> a little bit, starting from <span class="math-container">$0$</span>, then your utility also increases. Therefore your utility function cannot have a maximum at <span class="math-container">$0$</span>. Rather, its maximum must be attained at some positive <span class="math-container">$x$</span>.</p>

CC BY-SA 4.0
2023-02-28T08:05:20.240Z
2023-02-28T08:05:20.240Z
VARulle
<p>I couldn't get my head around this part. Basically, I have to prove that a consumer has to hold a positive amount of assets, i.e. <span class="math-container">$x &gt; 0$</span>.</p>
<p>A hint suggested to find take the FOC, and then set <span class="math-container">$x = 0$</span> and I would see that FOC is greater than 0, meaning that <span class="math-container">$x = 0$</span> cannot possibly be a utility maximizing choice, and the consumer must hold a positive amount of assets.</p>
<p>What does FOC being greater than 0 have anything to do with <span class="math-container">$x = 0$</span> not being the a utility maximizing choice?</p>

<consumer-theory><self-study>
FOC greater than 0
kyrie3
# FOC greater than 0

(Asked by: username_0 on 2023-02-28)

I couldn't get my head around this part. Basically, I have to prove that a consumer has to hold a positive amount of assets, i.e. $x > 0$.

A hint suggested to find take the FOC, and then set $x = 0$ and I would see that FOC is greater than 0, meaning that $x = 0$ cannot possibly be a utility maximizing choice, and the consumer must hold a positive amount of assets.

What does FOC being greater than 0 have anything to do with $x = 0$ not being the a utility maximizing choice?

**Tags:** consumer-theory, self-study


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## 3 answers

### Answer by username_2 on 2023-02-28. Score: 4 | Accepted answer

It's not quite possible to have "FOC greater than 0", since the FOC is a *condition*, not a number. What is really meant here is the derivative of the utility function.

If your utility function is given by $u(x)$, then the FOC for a maximum would be $u'(x)=0$. Now if you calculate the derivative $u'(x)$ and then substitute $0$ for $x$, then you get the derivative at $0$, written $u'(0)$.

If it turns out that $u'(0)>0$, then this means that your utility function is *locally increasing* at $0$. Hence, if you increase $x$ a little bit, starting from $0$, then your utility also increases. Therefore your utility function cannot have a maximum at $0$. Rather, its maximum must be attained at some positive $x$.


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### Answer by username_1 on 2023-02-28. Score: 1

Remember what is the **goal of optimization**: It is to find some maximum or minimum of a function. Potentially, to find maximum or minimum such that something must be satisfied (constraints)

First order conditions tell you that the directional slopes (partial derivations) of function are $0$. If you think about it, slope being equal to $0$ indicates local maximum or minimum (the function either stops increasing or decreasing).

Now, usually the result of taking FOC of some problem results in **functions** (or **conditions** that must hold for a choice to be optimum), let's call them $f'(\boldsymbol{x}) = 0$. If you plug concrete numbers into $\boldsymbol{x}$, for example $x = 0$, and with that your $f'(\boldsymbol{x}) \neq 0$, you know that the condition resulting from FOC **cannot be satisfied by that concrete** $\boldsymbol{x}$!

Only such choice $\boldsymbol{x}$ for which it holds $f'(\boldsymbol{x}) = 0$ can be optimal!

> **Comments:**
> **username_2**: This is not really true since it ignores the possibility of a boundary maximum. $f$ could be strictly decreasing.  
> **username_1**: Could you elaborate further? When doing Lagrangian (or KKT when necessary), even boundary maximum requires that $f'(\boldsymbol{x}) = 0$, where $f$ is the Lagrange function.  
> **username_2**: The $f$ you describe in your answer is defined on "numbers" $x$ and has a derivative denoted by the prime. This cannot be a Lagrange function.  
> **username_1**: It's more of a notation thing (by the way, the $x$ is bold in my text). If you read the text, you can see that I speak of conditions or functions and I just denote them by $f'()$. The $\boldsymbol{x}$  is vector of variables, not single "numbers", that is why it is bold. What concerns the rest, the $f'(\boldsymbol{x})$ is a vector of FOC. Just a reminder that I consider the multiplicator to be variable as well.  
> **username_2**: O.k., but then the prime in $f'({\bf x}) \ne 0$ has to be interpreted as a gradient and the $0$ as the zero vector. And this while the problem of the OP is obviously 1-dimensional...  

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### Answer by username_3 on 2023-02-28. Score: 0

If the question asks you to prove $x>0$, then the easiest you can do is to come up with the contradiction. Take $x=0$ and some $x>0$, show that the utility $u(x>0) > u(x=0)$. You are done.

Analytically, you could investigate your utility function. If it is increasing monotonically in $x$, then $x$ is going to hit the maximum possible constraint, unless you have a constraint that imposes exogenously $x=0$.


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