*n*should not decrease too fast compared to

*p.*But the lowest rate of change would be when the

*n*begins to converge on zero.

- Thread starter AmalR
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I tried to maximise n becauseYou speak of 'n'. That is not the question. Your task is to maximize 'I'.I= pn

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\(\displaystyle I=pn\)

We want \(I\) purely as a function of \(n\), and fortunately we are given a relationship between \(n\) and \(p\). Can you solve that for \(p\), and then substitute for \(p\) in the definition of income, so that you have income as a function of \(n\)?

Once you do that, then you can find the extrema via differentiation.

Hi, thanks! I decided to use Mark's method to putYou want to maximize I

Now I = pn = p*85e^{-.01p}.

Now calculate I'(p) and find the p that maximizes I.

Continue from here.

Thanks so much! I realise that I'm supposed to differentiate this new function to find the maximum where \(\displaystyle dI/dn=0\), and then reject any solutions with a -ve

\(\displaystyle I=pn\)

We want \(I\) purely as a function of \(n\), and fortunately we are given a relationship between \(n\) and \(p\). Can you solve that for \(p\), and then substitute for \(p\) in the definition of income, so that you have income as a function of \(n\)?

Once you do that, then you can find the extrema via differentiation.

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I think that my method is a bit easier, at it looks like that on the surface. Instead of solving for n, you just have to put the pHi, thanks! I decided to use Mark's method to putIas a function of n instead and then differentiated this function to find the stationary point whereIwould be at its maximum. Thank you.

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That is fine, but what do you do with the n's that survived? Are they max value, min values or point of inflections?Thanks so much! I realise that I'm supposed to differentiate this new function to find the maximum where \(\displaystyle dI/dn=0\), and then reject any solutions with a -venor any other impossible circumstance. Thanks, again!

I'm not sure if any of them match the description since the equation of the derivative is still exponential (derivative of \(\displaystyle e^x\) is still \(\displaystyle e^x\)). I subbed in 0 for \(\displaystyle dI/dn\) and I only got one result (I assume because there is only one x-intercept). Final answer was roughly 31, thanks so much for your help!That is fine, but what do you do with the n's that survived? Are they max value, min values or point of inflections?