For the demand function q=D(p)=500/(p+6)^5, find the following.

Question

For the demand function q=D(p)=500/(p+6)^5, find the following,

a) The equation for elasticity. (This is where I need the most help. Every step is extremely helpful.)

b) The elasticity at p=4, stating whether the demand is elastic, inelastic or has unit elasticity

c) The value(s) of p for which total revenue is a maximum (assume that p is in dollars)

Answer 1

E = - dq/dp/(q/p)

q = 500/(p+6)^5

a)dq/dp = -2500/(p+6)^6

Then, - dq/dp/(q/p) = -2500/(p+6)^6/(500/(p+6)^5/p) =

5p/(p+6)

b) At p = 4, 5p/(p+6) = 5*4/(4+6) = 20/10 = 2

c) p is a maximum when the elasticity equals 1.

5p/(p+6) = 1

5p = p+6

4p = 6

p = 3/2 = $1.50.

Note that E = 5p/(p+6) = 5(p+6-6)/(p+6) = 5 - 6/(p+6). Thus, this will be less than 1 for p < 3/2, and p = 3/2 is a maximum.

$1.50