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