The net monthly profit, in dollars, from the sale of a certain item is given by the formula P(x) = 10^6[1 + (x-1)e^0.001x], where x is the number ofitems sold. Determine the number of items that...

The net monthly profit, in dollars, from the sale of a certain item is given by the formula P(x) = 10^6[1 + (x-1)e^0.001x], where x is the number of

items sold. Determine the number of items that yield the maximum profit. At full capacity, the factory can produce 2000 items per month.

Expert Answers
beckden eNotes educator| Certified Educator

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

When in doubt take the derivative using the product property.

`P'(x)=10^6((x-1)(0.001e^(0.001x))+e^(0.001x))=10^6e^(0.001x)(0.001x+0.999)`

P'(x)=0 when (0.001x+0.999)=0

Solving for x =-999.  This is the only critical points, but it is not possible to make -999 items.

We also have to check x=0 and x=2000

`P(0)=10^6(1+0e^(0.001x))=10^6`

`P(2000)=10^6(1+1999e^(0.001(2000))=10^6(1+1999e^2)~~1.477xx10^10`

` `The solution is to produce 2000 items.