# Was told to use derivatives but not sure how that would help

#### AmalR

##### New member
I understand how to do a derivative of an exponential function and that this is an exponential decay. I can also see that the rate of change would probably have to be low because 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.

#### tkhunny

##### Moderator
Staff member
You speak of 'n'. That is not the question. Your task is to maximize 'I'. I = pn

#### AmalR

##### New member
You speak of 'n'. That is not the question. Your task is to maximize 'I'. I = pn
I tried to maximise n because I would increase as a result. But I found that when n increases past a certain point, p becomes very small. I just don't know how to relate this to derivatives?

#### MarkFL

##### Super Moderator
Staff member
We are asked to maximize income, and we are given:

$$\displaystyle I=pn$$

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.

#### Jomo

##### Elite Member
You want to maximize I

Now I = pn = p*85e-.01p.
Now calculate I'(p) and find the p that maximizes I.
Continue from here.

#### AmalR

##### New member
You want to maximize I

Now I = pn = p*85e-.01p.
Now calculate I'(p) and find the p that maximizes I.
Continue from here.
Hi, thanks! I decided to use Mark's method to put I as a function of n instead and then differentiated this function to find the stationary point where I would be at its maximum. Thank you.

#### AmalR

##### New member
We are asked to maximize income, and we are given:

$$\displaystyle I=pn$$

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.
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 n or any other impossible circumstance. Thanks, again!

#### Jomo

##### Elite Member
Hi, thanks! I decided to use Mark's method to put I as a function of n instead and then differentiated this function to find the stationary point where I would be at its maximum. Thank you.
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 pmax value into the n formula. Try it both ways and see for yourself if I am right or wrong.

#### Jomo

##### Elite Member
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 n or any other impossible circumstance. Thanks, again!
That is fine, but what do you do with the n's that survived? Are they max value, min values or point of inflections?

#### AmalR

##### New member
That is fine, but what do you do with the n's that survived? Are they max value, min values or point of inflections?
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!