Stats Solutions

(See Steps) Previously, the concept of price elasticity of demand was examined in situations where a product’s Demand model was linear . In some instances,

Question: Previously, the concept of price elasticity of demand was examined in situations where a product’s Demand model was linear . In some instances, this linear Demand model provided the basis to develop the product’s Revenue model. Not surprisingly, a 2nd order Revenue model would result from a 1st order Demand model.

Now that 3rd order models have been examined, it should be possible to deal with price elasticity of demand situations where a product’s Demand model is quadratic , and where the development of a Revenue model from this 2nd order Demand model results in a 3rd order model. For example, given the quadratic Demand model x = 187,500 – 25( p )2 , the Revenue model would be R = ( unit price )( units sold )= ( p )( x ) = [( p )][ 187,500 – 25( p )2 ] ==> 187,500( p )1 – 25( p )3.

It should be possible to find that the unit price " p " associated with maximum Revenue (assuming no limitations from restricted domain considerations on values of " p "). Thus…for Revenue models in the form R = f ( p ), " R max" occurs at the unit price " p " associated with [ d ( R )/ d ( p )] = ( R )' = 0 .

R = 187,500( p )1 – 25( p )3 ==> ( R )' = 187,500 – 75( p )2

@ 0 = ( R )' = 187,500 – 75( p )2 ==> 75( p )2 = 187,500 ==>

1( p )2 = $ 2,500.00 ==> 1( p ) = $ 50.00 [ignore, of course, the negative root –$50.00]

@ p = $ 50 … R max = 187,500( 50 )1 – 25( 50 )3 = $ 6,250,000 .

Previously, it was demonstrated that the price " p " where maximum Revenue occurs (assuming no limitations from restricted domain considerations on values of " p ") is the same as the price " p " where unit elasticity ( ɳ = –1) occurs. Given that the general form of any product’s elasticity expression may be written as…

ɳ = [( p ) / ( x )] [ d ( x ) / d ( p )] or [( p ) / ( x )] [( x )′]