As we know that $Q*P=const.$ for Cobb-Douglas preferences, we can thus conclude that $\frac{dQ/Q}{dP/P}$ is always $-1$:
$$ QP=const. \implies 0=d(PQ)=Q\ dP+P\ dQ \implies \frac{dQ}{Q}=-\frac{dP}{P} $$
$\begingroup$
$\endgroup$
0
Add a comment
|
$\begingroup$
$\endgroup$
7
This will be a weird answer:
As we know that $Q*P=const.$ for Cobb-Douglas preferences.
$$ QP=const. \implies 0=d(PQ)=Q\ dP+P\ dQ \implies \frac{dQ}{Q}=-\frac{dP}{P} $$ thus we can conclude that $\frac{dQ/Q}{dP/P}$ is always $-1$.
-
$\begingroup$ @j3141592653589793238 I still don't know what you were asking, I copied the answer from the body of your question. $\endgroup$ – Giskard May 12 at 10:35
-
$\begingroup$ I'm asking whether the assumption about Cobb-Douglas preferences always leading to iso-elastic demand curves is correct. Should I avoid such yes/no questions? Sorry. I just wasn't quite sure. @Giskard $\endgroup$ – j3141592653589793238 May 12 at 10:35
-
1$\begingroup$ Allright. If you feel my copying the body of your question into an answer helped, you are welcome. $\endgroup$ – Giskard May 12 at 10:40
-
1$\begingroup$ Sure, just needed some confirmation from a professional! :) @Giskard $\endgroup$ – j3141592653589793238 May 12 at 10:41
-
2
$\begingroup$
$\endgroup$
1
Yes.
I have to include at least 30 characters in an answer, so let me repeat: Yes.
-
$\begingroup$ Cheers. I feel very much confirmed now indeed. $\endgroup$ – j3141592653589793238 May 12 at 16:58
