<p>If you look at the case <span class="math-container">$x_1=x_2=\ldots x_n=t&gt;0$</span>, you have
<span class="math-container">$$F(x_1,x_2,\ldots,x_n)=t^{a_1}\cdot t^{a_2}\cdots t^{a_n}.$$</span>
Verify that the function <span class="math-container">$t\mapsto t^{a_1}\cdot t^{a_2}\cdots t^{a_n}$</span> is not concave. This implies that <span class="math-container">$F$</span> is not concave either.</p>

CC BY-SA 4.0
2021-03-01T21:54:36.760Z
Alternatively, recognize that the way the question is formulated implies that the answer doesn't depend on $n$. Then choose $n=1$ for simplicity and the answer is obvious. (Or, if not, calculate the second derivative.)
VARulle
CC BY-SA 4.0
2021-03-01T14:42:29.617Z
2021-03-01T14:42:29.617Z
Michael Greinecker
<p>Given the function <span class="math-container">$F(\mathbf{x})=x^{a_1}_1x^{a_2}_2 \ldots x^{a_n}_n$</span> defined on the set <span class="math-container">$S=\{\mathbf{x}=(x_1, \ldots, x_n) \in \mathbb{R}^n: x_1&gt;0, \ldots ,x_n&gt;0\}$</span> with <span class="math-container">$a_1,a_2,\ldots,a_n &gt; 0$</span> and <span class="math-container">$a_1+a_2+\ldots+a_n=3$</span>, I want to:</p>
<p>(i) Show that <span class="math-container">$\mathbf{x} \cdot \nabla F(\mathbf{x}) = 3F(\mathbf{x})$</span> at every <span class="math-container">$\mathbf{x}$</span>, where <span class="math-container">$\nabla F(\mathbf{x}) = (\frac{\delta F(\mathbf{x})}{\delta x_1},\ldots,\frac{\delta F(\mathbf{x})}{\delta x_n})$</span>.</p>
<p>I was able to work out that <span class="math-container">$\mathbf{x} \cdot \nabla F(\mathbf{x}) = a_1x^{a_1}_1 + \ldots + a_nx^{a_n}_n$</span> but got no further. Need help with this part!</p>
<p>(ii) Determine whether <span class="math-container">$F(\mathbf{x})$</span> is concave in <span class="math-container">$\mathbf{x}$</span> on the set <span class="math-container">$\mathbf{x}$</span>.</p>
<p>My first thought was to use a Hessian matrix but that would be too tedious for this function. Is there a better method?</p>

CC BY-SA 4.0
2021-03-01T14:21:48.530Z
To check whether $F$ is concave or not, you could take a look at the case $x_1=x_2=\ldots=x_n=t>0$. Then you get a function of a single argument $t$, and this will turn out to be enough for checking whether $F$ is concave on $S$ (not on $x$!) or not.
Michael Greinecker
CC BY-SA 4.0
2021-03-01T14:29:01.113Z
@MichaelGreinecker consider expanding your comment into an answer
1muflon1
CC BY-SA 4.0
2021-03-01T14:42:50.770Z
@1muflon1 Done.
Michael Greinecker
CC BY-SA 4.0
2021-03-01T21:49:51.147Z
Check $\mathbf{x} \cdot \nabla F(\mathbf{x})$ again, you seem to have "differentiated away" the constants in a product, which gave you the wrong result.
VARulle
<mathematical-economics>
Proof of Concavity of Cobb-Douglas Function
kékszajkók
# Proof of Concavity of Cobb-Douglas Function

(Asked by: username_0 on 2021-03-01)

Given the function $F(\mathbf{x})=x^{a\_1}\_1x^{a\_2}\_2 \ldots x^{a\_n}\_n$ defined on the set $S=\{\mathbf{x}=(x\_1, \ldots, x\_n) \in \mathbb{R}^n: x\_1>0, \ldots ,x\_n>0\}$ with $a\_1,a\_2,\ldots,a\_n > 0$ and $a\_1+a\_2+\ldots+a\_n=3$, I want to:

(i) Show that $\mathbf{x} \cdot \nabla F(\mathbf{x}) = 3F(\mathbf{x})$ at every $\mathbf{x}$, where $\nabla F(\mathbf{x}) = (\frac{\delta F(\mathbf{x})}{\delta x\_1},\ldots,\frac{\delta F(\mathbf{x})}{\delta x\_n})$.

I was able to work out that $\mathbf{x} \cdot \nabla F(\mathbf{x}) = a\_1x^{a\_1}\_1 + \ldots + a\_nx^{a\_n}\_n$ but got no further. Need help with this part!

(ii) Determine whether $F(\mathbf{x})$ is concave in $\mathbf{x}$ on the set $\mathbf{x}$.

My first thought was to use a Hessian matrix but that would be too tedious for this function. Is there a better method?

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: To check whether $F$ is concave or not, you could take a look at the case $x_1=x_2=\ldots=x_n=t>0$. Then you get a function of a single argument $t$, and this will turn out to be enough for checking whether $F$ is concave on $S$ (not on $x$!) or not.  
> **username_2**: @MichaelGreinecker consider expanding your comment into an answer  
> **username_1**: @1muflon1 Done.  
> **username_3**: Check $\mathbf{x} \cdot \nabla F(\mathbf{x})$ again, you seem to have "differentiated away" the constants in a product, which gave you the wrong result.  

---

## 1 answers

### Answer by username_1 on 2021-03-01. Score: 4

If you look at the case $x\_1=x\_2=\ldots x\_n=t>0$, you have
$$F(x\_1,x\_2,\ldots,x\_n)=t^{a\_1}\cdot t^{a\_2}\cdots t^{a\_n}.$$
Verify that the function $t\mapsto t^{a\_1}\cdot t^{a\_2}\cdots t^{a\_n}$ is not concave. This implies that $F$ is not concave either.

> **Comments:**
> **username_3**: Alternatively, recognize that the way the question is formulated implies that the answer doesn't depend on $n$. Then choose $n=1$ for simplicity and the answer is obvious. (Or, if not, calculate the second derivative.)  

---

