# Thread: Price Elasticity with Differentiation

1. ## Price Elasticity with Differentiation

I'm having trouble with this problem from a review sheet of mine.

Problem 2
Consider the price-demand equation x=f(p)=(144-2p)^(1/2) , 0>p>72.

a) Calculate the elasticity of demand at p 20, 40 and 60. Describe each case as
elastic or inelastic. For what value of p is demand unit elastic?

b) Express total revenue as a function of price p and find the value of p which
maximizes total revenue.

c) Explain, for the given price-demand equation, the connection between elasticity of
demand and the increasing/decreasing behaviour of the total revenue function
depending on price p !

What I'm really struggling with is converting the equation to the correct one for a). The answer given by the review is E=(p)/(144-2p). But the closest I've come was E=1/(144-2p)

Any help is great!

2. ## Re: Price Elasticity with Differentiation

the price elasticity of demand is defined as $\displaystyle \epsilon = \frac{dx}{dP} \frac{P}{x}$

Evaluating the derivative:
$\displaystyle \epsilon = \left( \frac{-1}{(144-2p)^{0.5}} \right) \frac{p}{x}$

This is technically an answer, but it isn't much help as we have two unknowns (x,p) on the right. However you are given the demand function, so you know the relationship between x and p. subsitute $\displaystyle x = (144-2p)^{0.5}$

$\displaystyle \epsilon = \left( \frac{-1}{(144-2p)^{0.5}} \right) \frac{p}{(144-2p)^{0.5}}$

which simplifies to:

$\displaystyle \epsilon = \frac{-p}{(144-2p)}$

i assume this is what your text book says (remember own price elasticity is almost always negative)