Shark Inc. has determined that demand for its newest netbook model is given by lnq−3lnp+0.004p=7lnq−3lnp+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:
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:
