a) Differentiate both sides of lnq − 3lnp + 0.003p=7 with respect to p, keeping in mind that q is a function of p and so using the Chain Rule to differentiate any functions of q:
(1/q)(dq/dp) − 3/p + 0.003 = 0
dq/dp = (3/p − 0.003)q.
So E(p) = dq/dp (p/q) = (3/p − 0.003)(q)(p/q) = (3/p − 0.003)p = 3 − 0.003p.
b) The revenue is pq.
Note that (d/dp) of pq = q + p dq/dp = q[1 + dq/dp (p/q)] = q(1 + E(p)), which is zero when E(p) = −1. Therefore, to maximize revenue, set E(p) = −1:
3 − 0.003p = −1
0.003p = 4
p = 4/0.003 = 4000/3 = 1333.33
