Hi everyone just wondering if someone could give me a push in the right direction for a problem I'm working on. I've been given the statement:

P(x)=10^6(1+(x-1)e^-0.001x)

Where P is profit and x is the number of items being produced. It wants to know what the max profit is when the domain is [0,2000]. Now I've concluded to do so I must take the derivative of this.Yet I'm not exactly sure how to go about doing that. So if anyone could purhaps break that down into a step by step format for me it would be much appreciated. The answer is given at the back of the book as x=1001

Thank you in advance

After some more time looking at it I can see where my derivative went wrong. For those interested I will post my solution.

10^6[1+(e^-0.001x)(x-1)] (rearranged)

P'(x)= 10^6[(e^-0.001x) (1) + (x-1)(e^-0.001x)(-0.001)] (Use the product and chain rules)

=10^6[(e^-0.001x) +(-0.001xe^-0.001x)(x-1)] (simplifying step)

=10^6[(e^-0.001x) +(0.001e^-0.001)(-0.001xe^-0.001)] (we can now remove any common factors in this case e^-0.001)

=(10^6)(e^-0.001)[1.001 - .001x] (run the 10^6 through the brackets)

=(e^-0.001)[1.001x10^6 -1000x] (To find the critical points we set this =0 and since (e^-0.001) can never be 0 we ignore it)

0=[1.001x10^6-1000x] (solve for x)

1.001x10^6=1000x

x=1001

now plug 1001 into the original equation to get your answer. Also you must do the domain points of 0 and 2000 to ensure you have the correct answer

so P(0)=0 P(1001)= 3.685x10^9 P(2000)=2.71x10^9

Thank you to all who took the time to read and try an puzzle this out it is much appreciated