# Thread: Elasticity of Demand problem

1. ## Elasticity of Demand problem

E (p) = -f'(p) x P / f (p)

The question says: Compute the elasticity of demand and determien whether it is elastic (>1) , inelastic (<1) or unitary (1) at the given unit price p0. if the unit price is increased slightly from po, will revenue increase or decrease?

1. x = -4/3p + 24; p0 = 6

I'm not sure how to plug this in into the formula. I tried a few ways out but coudln't get it to work. Thanks for the help in advance.

2. Originally Posted by agent2421

E (p) = -f'(p) x P / f (p)

The question says: Compute the elasticity of demand and determien whether it is elastic (>1) , inelastic (<1) or unitary (1) at the given unit price p0. if the unit price is increased slightly from po, will revenue increase or decrease?

1. x = -4/3p + 24; p0 = 6

I'm not sure how to plug this in into the formula. I tried a few ways out but coudln't get it to work. Thanks for the help in advance.
Hi agent2421,

$\displaystyle f(p)=-\frac{4}{3p}+24\Rightarrow{f(6)=23.78}$

$\displaystyle f'(p)=\frac{4}{3p^2}\Rightarrow{f'(6)=0.0370}$

Therefore, $\displaystyle E(p)=-p\frac{f'(p)}{f(p)}=-6\times\frac{0.0370}{23.78}=-9.35\times{10^{3}}$

3. Thanks for the help first of all... unfortunately i don't think that is the right answer. The answe it gives in my book is:

E (p) = p/18-p; E (6) = 1/2 < 1 , therefore it is inelastic... But i have no idea how to get that.

4. Originally Posted by agent2421
Thanks for the help first of all... unfortunately i don't think that is the right answer. The answe it gives in my book is:

E (p) = p/18-p; E (6) = 1/2 < 1 , therefore it is inelastic... But i have no idea how to get that.
Well, in your first post the you have written the function as, f(p) = -4/3p + 24. So I thought it is $\displaystyle f(p)=-\frac{4}{3p}+24$. But according the correct answer it should be $\displaystyle f(p)=-\frac{4}{3}p+24$. If you do not know how to use latex you should have written it as f(p) = -4p/3 + 24 or f(p) = (-4/3)p + 24. I hope this clears everything.