Hard logarithmic/implicit differentiation question.

[Senjai]

QUESTION HAS BEEN SOLVED ! :) THANKS MARK!

Homework Statement

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, 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 $$p \geq 100$$. 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

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:

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

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

Isolating i got:

$$ln(q) = 3ln(p) - 0.002p +7$$

$$q = e^\left(3ln(p) - 0.002p +7\right)$$

Then I differentiated using the chain and logarithm rules

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

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

I got

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

But this is wrong, why?

 

[Mark44]

Homework Statement

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, 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 $$p \geq 100$$. 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

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:

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

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

Isolating i got:

$$ln(q) = 3ln(p) - 0.002p +7$$

$$q = e^\left(3ln(p) - 0.002p +7\right)$$

This looks fine. You might want to simplify it, though, to e⁷*p³*e^-0.002p.

Then I differentiated using the chain and logarithm rules

Instead of doing it this way, why don't you differentiate the original equation implicitly. I think that would be simpler. When you're ready to put your answer for E(p), you can replace q using the formula you got above, so that E(p) is written in terms of p alone.

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

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

I got

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

But this is wrong, why?

 

[Senjai]

is q' wrong?

I did it implicitly using the original equation and got the same answer.

Implicit

$$(ln(q)-3ln(p)-0.002p)' = (7)'$$

=

$$\frac{q'}{q} - \frac{3}{p} - 0.002 = 0$$

And rearranged to get the same answer as my previous q' after subbing in my original q. (multiplying both sides by q.)

 

[Mark44]

You have a sign error.
From your last equation in the previous post,
q'/q = 3/p + 0.002 ==> q' = q(3/p + 0.002)

 

[Senjai]

According to [PLAIN]https://webwork.elearning.ubc.ca/webwork2_files/tmp/equations/80/fa9842dd3479ee1baaab655e02a1c41.png

Isnt it q'/q = 3/p - 0.002 ??

Thanks for your assistance so far by the way.

Edit: oh from the post I just made, yes i see that, but the answer is still wrong (online submission, unlimited tries) the last post is incorrect, the original post is it's correct form

 

[Mark44]

I was going by what you had in post #3 which I didn't notice was different from post #1.

What you have in post #2 should be simplified.
$$E\left(p\right) = -1 * \left[\left(e^\left(3ln(p) - 0.002p +7\right)\right) * \left(\frac{3}{p} - 0.002\right)\right] * \frac{p}{e^\left(3ln(p) - 0.002p +7\right)}$$
Cancel the e^{3lnp - .002p + 7} factors in the numerator and denominator.

 

[Senjai]

My god, that's all I had to do, after that I had the correct answer.

Thank you for your help mark :) Must be tired or something.