Shark Inc. has determined that demand for its newest netbook model is given by lnq-3lnp+0.004p=7ln⁡q-3ln⁡p+0.004p=7, where q is the number of netbooks Shark can sell at a price of pp dollars per unit. Shark has determined that this model is valid for prices p≥100p≥100. You may find it useful in this problem to know that elasticity of demand is determined to be E (p) = dqdppqE (p) = dqdppq

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