Shark Inc. has determined that demand for its newest netbook model is given by lnq−3lnp+0.005p=7, where q is the number of netbooks Shark can sell at a price of p dollars per unit. Shark has determined that this model is valid for prices p≥100. You may find it useful in this problem to know that elasticity of demand is defined to be E(p)=dq/dp (p/q) a)Find E(p) Your answer should only be in terms of p b)What price will maximize revenue.
It's actually lnq-3lnp+0.005p=7
Are you looking for the derivative method?
I tried getting q'(p/q) and I got 3-0.005p but that was wrong
my q'=((3/p)-0.005)q
Hmmm... well, did you do implicit differentiation because you have both variables in there?
yup
I don't know what I'm doing wrong
Elasticity of demand is just dq/dp. Where is the other p/q factor from?
I'm having so much trouble finding elasticity though.. I keep gettng 3-0.005p
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