# Proof of Concavity of Cobb-Douglas Function

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?

• 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 Mar 1 at 14:21
• @MichaelGreinecker consider expanding your comment into an answer – 1muflon1 Mar 1 at 14:29
• @1muflon1 Done. – Michael Greinecker Mar 1 at 14:42
• 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 Mar 1 at 21:49

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.
• 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 Mar 1 at 21:54