# Cobb-Douglas utility function

A simple Cobb-Douglas utility function:

$$U(X,Y)=\frac{X^{(1-\alpha)}Y^{\alpha}}{(1-\alpha)^{(1-\alpha)}\alpha^{\alpha}}$$

Here, I don't understand why we need the denominator: $(1-\alpha)^{(1-\alpha)}\alpha^{\alpha}$. What is the meaning of this denominator term?

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## 1 Answer

It is a normalisation factor intended to make $U(1-\alpha,\alpha)=1$ and is not necessary for the definiton of Cobb-Douglas form. In fact, the most general way to define Cobb-Douglas production function is as follows:

$$U(x_1,x_2,\ldots, x_n)=\Pi_{i} {x_i}^{\lambda_i} \text{ where }\sum_i \lambda_i=1$$

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Is $U(1-\alpha,\alpha)=1$ works for every pair of $(X,Y)$? Suppose we have the constraint, $X+Y=1$. Then, $(X,Y)=(0.2,0.8)$, $U=0.68$. And $(X,Y)=(0.9,0.1)$, $U=0.73$, for $\alpha=0.4$. – Bill TP Feb 19 '13 at 0:29
No, you are normalising utilities of different proportions using $U(1-\alpha,\alpha)$. To understand why $U(1-\alpha,\alpha)$ acts as some benchmark, we will need more context and the details of where you came across this. – Bravo Feb 19 '13 at 0:31
I found that $U(X,Y)=1$ when $(X,Y)=(0.6,0.4)$ for $\alpha=0.6$. And $U(X,Y)=2$ when $(X,Y)=(1.2,0.8)$ for $\alpha=0.6$. Would you please interpret these results? – Bill TP Feb 19 '13 at 0:42
It means if you allocate 0.6 of your time on $X$ and 0.4 on $Y$, then your utility is linear in $(X,Y)$. – Bravo Feb 19 '13 at 0:49