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1. The problem statement, all variables and given/known data

Shark Inc. has determined that demand for its newest netbook model is given by [PLAIN]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/80/fa9842dd3479ee1baaab655e02a1c41.png, [Broken] 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 [tex]p \geq 100[/tex]. You may ﬁnd it useful in this problem to know that elasticity of demand is deﬁned to be [PLAIN]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/c9/9d605854cd629ac16035b6765c1d051.png [Broken]

1) Find E(p), your answer should be in terms of p only.

2) What price will maximize revenue, if the price is less than 100, write NA

2. The attempt at a solution

My answer for #1 (which was wrong) using the formula was:

[tex]-[(e^(3*ln(p)-0.002p+7)*(3/p-0.002))]*(\frac{p}{(e^(3*ln(p)-0.002p+7)})[/tex]

First i solved for q. As all answers must be in terms of p.

Isolating i got:

[tex] ln(q) = 3ln(p) - 0.002p +7 [/tex]

[tex] q = e^\left(3ln(p) - 0.002p +7\right) [/tex]

Then I differentiated using the chain and logarithm rules

[tex] q' = \left(e^\left(3ln(p) - 0.002p +7\right)\right) * \left(\frac{3}{p} - 0.002\right) [/tex]

So when I put these values into: [tex] E\left(p\right) = -q' * \frac{p}{q} [/tex]

I got

[tex] E\left(p\right) = -1 * \left[\left(e^\left(3ln(p) - 0.002p +7\right)\right) * \left(\frac{3}{p} - 0.002\right)\right] * \frac{p}{e^\left(3ln(p) - 0.002p +7\right)} [/tex]

But this is wrong, why?

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# Hard logarithmic/implicit differentiation question.

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